Question

A ball is thrown straight up with an initial speed of 36 m/s . Part A How much time does it take for the ball to reach the top of its trajectory?

Answer 1

Answer:3.67 s

Explanation:

Given

Ball is thrown with an initial velocity of $u=36\ m/s$

As the ball moves upward it experience a declaration due to gravity which will slow down the ball

using

$v=u+at$

where v=final velocity

u=initial velocity

a=acceleration

t=time

Here $v=0$

$0=36-9.8\times t$

$t=\frac{36}{9.8}$

$t=3.67\ s$