# Answer to Question #1023 in Mechanics | Relativity for sabeen

Question #1023

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s. Ignore the effects of air resistance. (a) How long until the ball reaches its highest point? (b) How high above the ground does the ball go?

Expert's answer

When the ball reaches the highest point his velocity equals to zero:

v=v

12 = 10t

t = 6/5 = 1.2 s

From the equation of the motion:

h =h

v=v

_{0}- gt = 012 = 10t

t = 6/5 = 1.2 s

From the equation of the motion:

h =h

_{0}+v_{0}t - gt^{2}/2 = 1.75 + 12 * 1.2 - 10* 1.2^{2}/2 = 8.95 m.Need a fast expert's response?

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