Calculate Force Applied: Y=1/2at^2 Solution

[fubag]

You and your lab partner devise a clever experiment to measure the force applied to a ball when it is thrown. You throw a 200-{g} ball vertically upward. A high-speed video image of the throw indicates your hand pushed on the ball for 0.32 {\rm s}. The ball rises to a maximum height of 7.23 {\rm m} above your hand.

Calculate the force applied.


I tried.

Y = 1/2at^2

so

7.23m = 1/2(9.81m/s/s)(t^2)

Solved for t = 1.2 seconds...


Now I know F = ma

so is it simply F = 0.2kg*9.81m/s/s?

Im not sure

 

[Doc Al]

I tried.

Y = 1/2at^2

so

7.23m = 1/2(9.81m/s/s)(t^2)

Solved for t = 1.2 seconds...

Now find the speed of the ball as it leaves your hand.

Now I know F = ma

so is it simply F = 0.2kg*9.81m/s/s?

No. That's just the weight of the ball.

Instead, figure out the ball's acceleration during the time that you were pushing on it. Then apply Newton's 2nd law to find the net force. (Note that your hand is not the only force acting on the ball.)

 

[fubag]

ok so there is a normal force as well? means F = m(a-g)?

then to find speed of the ball as it leaves my hand, how do I know final velocity? if I threw it, shouldn't initial velocity be 0?So to push the ball it took 0.32 seconds... F = ma...

Am I supposed to use work here?

I know that acceleration due to gravity is 9.8m/s/s

So to find acceleration of the ball, I just set ma = m(a-g)?

 

[Doc Al]

ok so there is a normal force as well? means F = m(a-g)?

Right. The force of the hand is the normal force.

then to find speed of the ball as it leaves my hand, how do I know final velocity? if I threw it, shouldn't initial velocity be 0?

Once the ball leaves your hand, it's a projectile. The final speed of that projectile motion is zero. But what's the initial speed (as it leaves your hand).

To find the acceleration of the ball during the time you were pushing on it, you will use a bit of kinematics. Once you find the speed at which the ball left your hand (using the height and projectile motion), you can find the acceleration by change in velocity over time.

 

[fubag]

ok...

if final velocity is zero in the x direction and initial velocity in y direction is 0. And if I have the acceleration due to gravity.

I am getting 23.52 m/s as initial velocity? Using V = Vo + at, using a = -9.8 m/sThen I just find acceleration of the ball by 23.52 m/s / 0.32s? = 73.5 m/s/s?

 

[Doc Al]

if final velocity is zero in the x direction and initial velocity in y direction is 0. And if I have the acceleration due to gravity.

Since it's thrown vertically, all we need to worry about is the y-direction. Let's call the initial velocity of the ball when it leaves the hand = Vo. The final velocity at the top of the motion (at y = 7.23 m) is zero.

I am getting 23.52 m/s as initial velocity? Using V = Vo + at, using a = -9.8 m/s

How did you get that number? What did you use for time?

 

[fubag]

ok I redid it and got the answer..

I just used

Vf^2 = Vi^2 +2ad, with a= -9.8, d = 7.32, and Vf = 0 at highest point..

i then used a = change in velocity/time

So I got the acceleration, then I used multiplied by mass, of 0.2 kg.

Worked.

Thanks!