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

Step-by-step 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'=((2u)^2)/(2g)`

`s'=4* (u^2)/(2g)`

`s'=4* 72`

`s'=288\ m`