Question

You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.1 s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward.
(a) Use the momentum principle to determine the speed that the ball had just AFTER it left your hand.
vinitial = ?? m/s

(b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.
h = ?? m

Answer 1

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

$v_f$ = Final velocity

$v_i$ = Initial velocity

$t_f$ = Final time

$t_i$ = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

$\Delta P=F\Delta t$

Force

$F=mg$

So,

$m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s$

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

$K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m$

The maximum height above your hand reached by the ball is 5.93505 m