You throw a 134 g ball straight up to a height of 43.0 m.
(a) Assume that the displacement of your hand is 0.70 m as you throw the ball straight up. How long is the ball is in your hand?
(b) Calculate the average force exerted by your hand while you are throwing it.
(a) release velocity = sqrt(2gH) = 29.0 m/s
Time in hand = 0.70 m/(Avg. Velocity) = 0.70/(29.0/2) = 0.048 s
(b) Force * Distance = Work = Potential Energy Gain at max. height
F*0.70 = M*g*H = 0.134*9.8*43 J
Solve for F in Newtons
To find the time the ball is in your hand (part a), we need to use the kinematic equation for displacement:
d = vit + (1/2)at^2
Where:
d = displacement
vi = initial velocity
a = acceleration
t = time
In this case, the ball is thrown straight up, so the initial velocity is the speed when the ball leaves your hand. The acceleration is the acceleration due to gravity, which is approximately -9.8 m/s^2 for objects near the Earth's surface.
Now let's calculate the initial velocity:
Given: Mass of the ball (m) = 134 g = 0.134 kg
Displacement of hand (d) = 0.70 m
According to the law of conservation of momentum, we can assume that the initial velocity of the ball is equal to the velocity of your hand when you release it.
So, we can use the equation for conservation of momentum:
mv = (m + M)V
Where:
m = mass of the ball
v = velocity of the ball
M = mass of your hand (assumed to be negligible)
V = velocity of your hand
Since we know the mass of the ball (m) and the displacement (d), but we need to find vi, we can rewrite the first equation in terms of vi:
d = vi × t + (1/2)at^2
Rearranging, we get:
vi = (d - (1/2)at^2) / t
Now we can substitute the known values:
vi = (0.70 m - (1/2)(-9.8 m/s^2)t^2) / t
To find the time the ball is in your hand (t), we can set up a quadratic equation by equating vi to 0 when the ball reaches its maximum height:
0 = (0.70 m - (1/2)(-9.8 m/s^2)t_max^2) / t_max
Simplifying and moving the terms around, we get:
0 = 0.70 m / t_max - (1/2)(-9.8 m/s^2)t_max
Now we can solve this quadratic equation to find the time t_max.