Question

A projectile is shot vertically upward with a given initial velocity. It reaches a maximum height of 72.0 m. On a second shot, the initial velocity is double, what now will be the maximum height that the projectile reaches?

Answer 1

Answer:

Explanation:

Given

maximum height reached $h=72\ m$

suppose Projectile is launched vertically with initial velocity u

applying equation of motion

$v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

Here final velocity is zero so

$s=frac{u^2}{2g}$

for twice initial velocity

$s'=\frac{(2u)^2}{2g}$

$s'=4\times \frac{u^2}{2g}$

$s'=4\times 72$

$s'=288\ m$