Question

A body is projected vertically upwards with a velocity 10m/s . It reaches maximum vertical height h in time t . In t/2 , the height covered is ???
1.)h/2
2)2h/5
3)3h/4
4)5h/8

Answer 1

3rd is the ans.........

Answer 2

Answer : Height covered is (c) 3h/4

Explanation :

It is given that,

Initial velocity of body, u = 10 m/s

It reaches at maximum height, v = 0

Acceleration due to gravity, $a=-g=-10\ m/s^2$

It reaches at height h in time t.

Using first equation of motion :

$v=u-gt$

$0=10-10t$

$t=1\ s$

To find h use third equation of motion :

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

$0-(10\ m/s)^2=2\times (-10\ m/s^2)\times h$

$h=5\ m$

If $t'=\dfrac{t}{2}=\dfrac{1}{2}\ s$, let H is the height.

Using second equation of motion :

$H=ut-\dfrac{1}{2}gt^2$

$H=10\ m/s\times \dfrac{1}{2}\ s-(10\ m/s^2)(\dfrac{1}{2}\ s)^2$

$H=\dfrac{15}{4}\ m$

or

$H=\dfrac{3h}{4}$

Hence, this is the required solution.