Answer to Question #259551 in Physics for Jai

Question

A boy tosses a coin upward with a velocity of +14.7 m/s. Find:

A. The maximum height reached by the coin.

B. Time of flight

C. Velocity when the coin returns to the hand.

D. Supposed the boy failed to catch the coin and the coin goes to the ground, with what velocity will it strike the ground? The boy's hand is 0.49 m above the ground.

Answer 1

A) We can find the maximum height reached by the coin from the kinematic equation:

$$v^2=v_0^2+2g(y_{max}-y_0),$$ $$0=v_0^2+2gy_{max},$$ $$y_{max}=\dfrac{v_0^2}{-2g}=\dfrac{(14.7\ \dfrac{m}{s})^2}{-2\times9.8\ \dfrac{m}{s^2}}=11\ m.$$

B) Let's first find the time the coin takes to reach maximum height:

$$v=v_0+gt,$$ $$0=v_0+gt,$$ $$t=\dfrac{v_0}{-g}=\dfrac{14.7\ \dfrac{m}{s}}{-(-9.8\ \dfrac{m}{s^2})}=1.5\ s.$$

Then, we can find the time of flight as follows:

$$t_{flight}=2t=2\times1.5\ s=3.0\ s.$$

C) We can find the velocity when the coin returns to the hand from the kinematic equation:

$$v=v_0+gt,$$ $$v=0+-9.8\ \dfrac{m}{s^2}\times1.5\ s=-14.7\ \dfrac{m}{s}.$$

The sign minus means that the velocity of the coin directed downward.

D) We can find the velocity of the coin when it hits the ground from the kinematic equation:

$$v^2=v_0^2+2g(y_{max}-y_0),$$ $$v^2=0+2g(0-y_0),$$ $$v=\sqrt{-2gy_0}=\sqrt{-2\times(-9.8\ \dfrac{m}{s^2})\times11.49\ m}=15\ \dfrac{m}{s}.$$