Answer to Question #116621 in Mechanics | Relativity for Adenagbe ayowole

Question

A volley ball is thrown straight upward on the moon with an initial speed of 30 m/s .Calculate: (a) The maximum height reached by the ball. (b) The time taken to reach that height. (c) Its velocity 20s after it is thrown. (d) When the ball’s height is 50m. (g = 1.60m/s2 on the moon).

Answer 1

Initial velocity of ball u = 30 $\frac{m}{sec}$

gravitational acceleration g = 1.60 $\frac{m}{sec^{2}}$

using equation motion

The maximum height reached by ball

$v^{2}-u^{2} = -2gh$

$0^{2}-30^{2} = - 2\times 1.60 \times h$ ( Final velocity v = 0, Negative sign is because of deceleration or oppose to gravity )

$h = 281.25 m$

The time taken to reach that height. 

$v = u - gt$

$0 = 30 -1.60\times t$

$t = 18.75 sec$

Its velocity 20s after it is thrown. 

$v = u - gt$

$v = 30 -1.60 \times 20$

$v = -2 \frac{m}{sec}$

When the ball’s height is 50m

$v^{2}-u^{2} = -2gh$

$v^{2}-30^{2} = -2\times 1.60 \times 50$

$v = 27.20 \frac{m}{sec}$

$0 = 30 -1.60\times t$

$v = 27.20 \frac{m}{sec}$