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

Answer 2

The speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.

What is momentum principal?

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the principal of momentum.

The net force using the momentum principle can be given as,

$F_{net}=\dfrac{\Delta P}{\Delta t}$

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. Therefore, the above equation for initial and final velocity and time can be written as,

$F_{net}=\dfrac{m(v_f-v_i)}{(t_f-t_i)}$

• (a) The speed that the ball had just AFTER it left your hand-

The mass of the ball is 1 kg, and it takes 2.2 s to go up and down. The net force acting on the body is mass times gravitational force.

Therefore, the above equation can be written as,

$mg=\dfrac{m(v_f-v_i)}{(t_f-t_i)}\\g=\dfrac{(v_f-v_i)}{(t_f-t_i)}$

As the final velocity is zero and initial time is also zero. Therefore,

$(-9.81)=\dfrac{(0-v_i)}{(1.1-0)}\\v_i=10.791 \rm m/s$

• (b) The maximum height above your hand reached by the ball-

Using the energy principle, we can equate the kinetic energy of the system to the potential energy of the system as,

$\dfrac{1}{2}mv_i^2=mgh\\\dfrac{1}{2}(10.791)^2=(9.81)h\\h=5.94\rm m$

Thus, the speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.

Learn more about the momentum here;

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