Question

When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high?

Answer 1

Answer:

Explanation:

Given

Initial speed is u=V

Maximum height of Pebble is H

Deriving maximum height of Pebble and considering motion in vertical direction

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

where v=final velocity

u=initial velocity

a=acceleration

s=Displacement

Final velocity will be zero at maximum height

$0-(V)^2=2\times (-g)\cdot H$

$H=\frac{V^2}{2g}$

i.e. maximum height is dependent on square of initial velocity

for twice the height

$2H=\frac{(V')^2}{2g}$

on comparing

$V'=\sqrt{2}V$