Question

A boy tosses a coin upward with a velocity of 14.7 m/s then catches it on his hand. Suppose 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

Given:

$v_0=14.7\:\rm m/s$

$h=0.49\:\rm m$

The law of conservation of energy says

$$\frac{mv_0^2}{2}+mgh=\frac{mv_1^2}{2}$$

Hence, the final speed of a coin

$$v_1=\sqrt{v_0^2+2gh}$$

$$=\sqrt{14.7^2+2*9.8*0.49}=15.0\:\rm m/s$$