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I’m 6 ft tall, wingspan is 6"5, Running Vertical is 31", I can easily palm the ball but when i go for the dunk i can get the ball to the rim but when the ball gets over it spills out and I end up grabbing the rim and a really close layup. Need help to find out much more to dunk with one hand, and how much does my vertical need to be to do tricks like 2 hands, 360, windmill, etc.
P.S. If anyone has tried a good vertical jump program that increased their vertical then just put that with your answers. Thanks

Just keep trying, you are close.

I’m in an intro physics class and am stumped on this question!

One method of pitching a softball is called the "windmill" delivery method, in which the pitcher’s arm rotates through approximately 360° in a vertical plane before the 198 gram ball is released at the lowest point of the circular motion. An experienced pitcher can throw a ball with a speed of 90.5 mi/h. Assume that the angular acceleration is uniform throughout the pitching motion, and take the distance between the softball and the shoulder joint to be 74.3 cm.
(a) Determine the angular speed of the arm in rev/s at the instant of release.
1 rev/s

(b) Find the value of the angular acceleration in rev/s2 and the radial and tangential acceleration of the ball just before it is released.
α 2 rev/s2
ac 3 m/s2
at 4 m/s2
(c) Determine the force exerted on the ball by the pitcher’s hand (both radial and tangential components) just before it is released.
Fr 5
Ft 6

Part (a)
Before I try to answer the question, let me first get a clear picture.
Is this what happens: Holding the ball stationary with one arm dropped at the
side of the body, the arm is rotated stretched out only once and releases the
ball at the lowest point. Is that correct? If it is, then I have the following things
to say, which I hope will help you.
The angular speed in revolutions per second (rev/s) is the number of
revolutions N that takes place in one second . It we denote the angular
speed by the Greek letter ω (omega) ,then
(1) ….. ω = ( N rev ) / ( 1 second) = N rev/s
Now, every time a ball is pitched, only one revolution takes place before
releasing the ball. That means N = 1 always. But the result is not always the
same. Some balls move rather slow, others move just fast enough, maybe,
but only a few move with extraordinary speed, like the one moving at
90 mph. They all have N = 1 revolution. But what makes them different? It’s
the time it takes to make that N = 1 revolution. The N = 1 revolution doesn’t
necessarily take place in 1 second if the angular speed is given in rev/s. You have
to find out how long it takes to make that N = 1 revolution. Now, in the case of a
ball thrown at a speed of 90.5 mph, how many seconds does it take to rotate the
arm stretched out only once? How do we answer that?
Think again about what happens. The faster the arm is whirled, the faster the ball
that comes out. The faster the ball moves, the farther the distance it travels in a
given fixed time. But is it only the ball that travels a certain distance? Is the ball
the only thing that moves in what we’re looking at? Of course not! Without
moving the arm, the ball will not move by itself. So there, the arm and the hand
also moved. But what distance did the arm travel? What distance did the hand
move? The distance traveled by the hand is none other than the length of the
circumference of the circle made by the outstretched arm. If we call that
circumference C, then
(2) ….. C = 2 * Pi * R where Pi = 3.1416
where R is the radius of the circle, which is also the distance of the hand from
the shoulder blade. But notice that R = 74.3 cm has a unit of distance different
from that of a ball moving in terms of miles. Before we can relate those two
distances, we have to make sure that they have the same unit of distance or
length. So, we have to make conversion of units before we can talk about other
things. We do the conversion as follows.
Using the SI system of measure we find the following:
………. 1 mile = 1.609 km = (1.609 km)(1000 m/km) = 1609 m
………. 1 hour = (1 hour)(60 min/hour)(60 seconds /min) =3600 second
………. 1 mph = 1 mile /(1 hour) = (1609 m)/(3600 s) = 0.4469 m/s
(3) ….. 90.5 mph = (90.5 mph)( 0.4469 m/s ) = 40.44 m/s
………. 1 m = 100 cm so there are 100cm/(1 m) = 100 cm/m
(4) ….. R = 74.3 cm = (74.3 cm)/(100 cm/m) = 0.743 m
Substituting (4) in (2) we get
(5) ….. C = (1.486)Pi m = 4.67 m
Notice that (5) and (3) now have the same unit of length in meter.
Going back to what we were discussing a little earlier, knowing the distance
in (5) that the hand moved in producing the throw given in (3), we can now
compute how long it took to rotate the arm just once. Since (5) is a distance
and (3) is speed, the time we’re looking for is just the distance in (5) divided
by the speed in (3). That is
(6) ….. T = [ (4.67 m ]/[ 40.44 m/s ] = 0.115 s
That’s how long it took to rotate the arm just once (N =1) in producing the
throw in (3). So, how do we get the angular speed now. By looking for the
number of revolutions N that will take place in one second , knowing that
1 rev takes place in 0.115 s. But how exactly do we do that. We do that by
using ratio and proportion. For a constant angular speed ω given in (1),
N changes with the same proportion as the change in ΔT. If N changes by
a factor K , so that the new value is now K * N, then ΔT also changes by
exactly the same factor, taking the new value K * ΔT. That means, the ratio
of N1 rev/ [ time ΔT1 ] must be the same as the ratio N2 rev/ [ time ΔT2 ]
Applying that to the case we’re discussing, we have
(7) ….. [ 1 rev / (0.115 s) ] = [ N rev / 1 s ]
Solving for N, we get N = 8.7 rev. That means 8.7 rev takes place in 1 second.
According to (1) above, that is the angular speed we’re looking for. At the
instant instant of release of a ball moving at speed of 90.5 mph, its angular
speed is 8.7 rev/s.
ANSWER : angular speed ω = 8.7 rev/s

Part (b)
Angular acceleration is defined as the change in angular speed Δω = ω2 – ω1
that takes place in time ΔT. If we denote the angular acceleration by the Greek
letter α (alpha), then
(8) ….. α = Δω / ΔT = (ω2 – ω1) / ΔT
Since ω1 = 0 at the start, when the ball is still at rest, and ω2 = 8.7 rev/s at
the instant of release of the ball after time ΔT = 0.115 s, we find t

The vertical leap test determines ultimatlety how athletic an athlete is and the average nba vertical leap is 28 inches. I’m 15 and my vertical leap is 31 inches( I’m not lying, if I was lying, this question would be pointless ). I’m just barely above the average NBA athlete in terms of athleticism. I have somewhat of a natural ability to control the ball. I not very skilled but I am very athletic and I have good handles. I’m 6′3 and 185 lbs. I can do tons of dunks. I can windmill and other crazy dunks. Since I’m only 15 and I have some handles( although overall I’m not very skilled) and I have a little more athletecism than the average nba player does that mean that with proper training and hard work I can at least be a good D1 balla?
By the way, Dwayne Wade’s vertical leap is 31.5 inches.

i think you’re ready. you should contact some European teams and make some money.

if all else fails the yankees will probably pay you too much money to do too little.

A softball pitcher has a windmill wind up in which a 0.25 kg ball moves on a vertical arc of radius r = 0.51 m. To accelerate the ball, she exerts a 28 N force parallel to the ball’s motion along the circular arc. The speed of the ball is 12 m/s at the top of the arc. With what speed is the ball released one half a revolution later at the bottom of the arc?

The radius r = 0.51 m

mass of ball m = 0.25 kg

Force on ball (F) = 28 N

Acceleration of ball due to force exerted by player (a) = F/m = 28/0.25
= 112 m/s^2

v (speed of ball) at top of arc (180 degrees) = 12 m/s

Therefore time it takes to go 180 degrees = v/a = 12 / 112
= 0.1071 s

Therefore time it takes for ball to make another 180 degrees before release (t) = 0.1071*2 s = 0.2142 s

a = 112 m/s^2

speed of ball at release = a*t = (112 * 0.2142) m/s = 24 m/s

There is a windmill, whose rotor can rotate freely in the vertical plane.
There is no frictional force.
In fact, let us assume that there is no contact force whatsoever
between the rotor and any other thing.
The rotor has four blades.
The rotor is completely at rest.
Two blades lie horizontally and the other two vertically.
The center of mass of the rotor is on the rotational axis.
The potential energy of the rotor is Mgh where h is the height
of the center of mass, and M is the total mass of the rotor.
The rotational inertia of the rotor is 4.3 kg m^2.
The length of each blade is 2.6 m.

1. the potential energy _____ constant.
2. the rotational kinetic energy _____ constant.
3. the angular momentum _____ constant.
4. the motion _____ a uniform circular motion.

1. isn’t
2. is
3. is
4. isn’t

A fastpitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 66 mph. The 0.19 kg ball is 53 cm from the pivot point at her shoulder.

A) Just before the ball leaves her hand, what is its centripetal acceleration?

B) At the lowest point of the circle the ball has reached its maximum speed. What is the magnitude of the force her hand exerts on the ball at this point?

C) At the lowest point of the circle the ball has reached its maximum speed. What is the direction of the force her hand exerts on the ball at this point?

Centripetal acceleration is V^2/R
V = 66 mph which has to be converted to km/h or m/s or whatever units you are interested in. I would use meters, etc since the are is 0.53 m and the ball is 0.19 kg.

At the lowest point, gravity is pulling down on the ball and her hand must compensate for this. It must also produce the force necessary to provide the centripetal acceleration for the ball to maintain the circular motion.

Centripetal force = m*V^2/R
Gravitational force = m*g

Force exerted by arm = m*V^2/R + m*g = m*[V^2/R + g]

At the lowest point gravity is acting downwards so the force of her hand will be directed upward. The force necessary to maintain an object in constant circular motion is directed perpendicular to the circumference of the circle.

There is a windmill, whose rotor can rotate freely in the vertical plane.
There is no frictional force.
In fact, let us assume that there is no contact force whatsoever
between the rotor and any other thing.
The rotor has four blades.
The rotor is completely at rest.
Two blades lie horizontally and the other two vertically.
The center of mass of the rotor is on the rotational axis.
The potential energy of the rotor is Mgh where h is the height
of the center of mass, and M is the total mass of the rotor.
The rotational inertia of the rotor is 4.3 kg m^2.
The length of each blade is 2.6 m.

81:c the kinetic energy _____ purely a rotational kinetic energy.
82:c the kinetic energy of r+m _____ constant.
83:c the motion of r+m _____ a circular motion.
84:c the motion of r+m _____ a uniform circular motion.
85:c the magnitude of the tangential accel. of the monkey _____ constant.
86:c the magnitude of the centripetal accel. of the monkey _____ constant.

87: The angular momentum of the monkey just before the landing is _____ J s.
Right after landing, the total angular momentum of r+m is also this value,
assuming that the time between "just before the landing" and
"just after the landing" is essentially zero.
[Since then, any angular momentum change of the total system is zero,
because d(angular momentum) = torque dt = 0, if dt = 0.]
88: The rotational inertia of r+m is _____ kg m^2.
89: The angular velocity of r+m just after the landing is _____ rad/s.

90: The total mechanical energy just after the landing is _____ J.
91: The total mechanical energy just before the landing is _____ J.
92:c Apparently, the total mechanical energy _____ conserved DURING
the landing, and that is quite OK.
The direction of the angular velocity of the monkey will not change at least
for the initial 180 degree rotation, independent of the monkey’s landing speed.
93:c This last statement is _____.
For the specific value of the monkey’s landing speed (and other parameters)
given above, the resulting motion will be a swinging back and forth
(an oscillation), instead of a going round and round.
94:c This last statement is _____.
The minimum amount of angle that the monkey needs to rotate before it
95: reaches its minimum speed is _____ degrees.

You should post these 1 at a time.

How can I increase my vertical leap? I want to be able to throw down a windmill by the end of next May. My height it 6′2" and my weight is 180 pounds. Also, right now I am able to do a two handed dunk just by taking only 3 steps. And my vert is approx around 38-41 inches. How many more inches is a windmill? Ive tried completing one 34 times once and I just couldn’t do it.

do calf raises and if you want u can even try programs such as air alert. If this is not good for you, the website below has great drills that will help u wit your vertical leap

A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 68mph . The 0.19kg ball is 51cm from the pivot point at her shoulder.
Just before the ball leaves her hand, what is its centripetal acceleration?
At the lowest point of the circle the ball has reached its maximum speed. What is the magnitude of the force her hand exerts on the ball at this point?
At the lowest point of the circle the ball has reached its maximum speed. What is the direction of the force her hand exerts on the ball at this point?

Centripetal acceleration is given by

a = v^2/R

so R = 0.51 m
and v = 68 mph = 30.4 m/s

a = 1812 m/s^2

Since just before the point of release there is no acceleration along the arc, the only acceleration is towards the centre (shoulder pivot).

So, F = ma

= 0.19 x 1812 = 344 N

I love basketball and I’m pretty good. Whenever I play in public I get extremely scared and I do bad. Becuase of that I couldnt make any basketball team but I’m very good. This fear has been a problem for a long time. This fear is irrational because I know I’m very good at basketball and very athletic. I have a 35 inch vertical leap and I can do a windmill slam dunk and I’m 6′3. I’m a good ball handler and a good shooter. Most people that I know had the same fear I do easily overcame it. Do you think I have a serious problem? Like a phobia. Most people that have seen me play basketball in private think I’m a prospect. What should I do? Should I see a psychalogist? I really think I should.

DUDE i have same problem. Am 5"9 and a guard but i have the same problem.