Question

A stone is thrown vertically upwards with a velocity 9.8 m/s.
a) What is the time taken to reach the maximum height?
b) What is the maximum height reached?
c) What is the velocity with which it reaches the ground?

Answer 1

Given,

Initial velocity = u = 9.8 m/s.

a) From the first equation of motion we have

$v=u+at$
$\Rightarrow 0=9.8-9.8t$
$\Rightarrow t=1s$
Time taken to reach the maximum height = 1 s
b) From the third equation of motion we have

$v^2-u^2=2as$
$\Rightarrow 0-(9.8{)}^2=2(-9.8)h$
$\Rightarrow h=4.9m$
Maximum height reached = 4.9 m
c) The stone will reach the ground with the same velocity with which it was thrown up i.e. 9.8 m/s.