Question

An object is thrown directly upwards from the ground at a velocity of 9ms. Recalling that the acceleration due to gravity is −gms2, find the time at which the object reaches its maximum height. What is this maximum height? Your answers should be in terms of g.

Answer 1

Answer:

Explanation:

Given

Object is thrown with a velocity of $u=9\ m/s$

Acceleration due to gravity is -g (i.e. acting downward)

Vertical distance traveled by object is given by

$v^2-u^2=2as$  

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

at maximum height final velocity is zero

$0-(9)^2=2\times (-g)\times (s)$

$s=\frac{81}{2g}=\frac{40.5}{g}\ m$

time taken to reach maximum height

using

v=u+at

0=9-gt

$t=\frac{9}{g}\ s$