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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?

Respuesta :

Answer:

Explanation:

Given

maximum height reached [tex]h=72\ m[/tex]

suppose Projectile is launched vertically with initial velocity u

applying equation of motion

[tex]v^2-u^2=2 as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

Here final velocity is zero so

[tex]s=frac{u^2}{2g}[/tex]

for twice initial velocity

[tex]s'=\frac{(2u)^2}{2g}[/tex]

[tex]s'=4\times \frac{u^2}{2g}[/tex]

[tex]s'=4\times 72[/tex]

[tex]s'=288\ m[/tex]

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