Question

a stone thrown vertically upwards with a speed of 5m/s attains a height h1.another stone thrown vertically from the same point with a speed of 10m/s attains a height h2 show that h2=4h1

Answer 1

Use formula v^2 = u^2 + 2aS. and solve it for both condition.and you will get your answer

Answer 2

Answer:

$h_{2}=4h_{1}$

Explanation:

Given that,

Speed of first stone = 5 m/s

Speed of second  stone = 10 m/s

Using equation of motion

$v^2 =u^2+2gs$

Where, v = final velocity

u = initial velocity

s = height

g = acceleration due to gravity

For first stone

$5^2=0+2\times9.8\times h_{1}$.....(I)

For second stone

$10^2=0+2\times9.8\times h_{2}$....(II)

Divided equation (I) by (II)

$\dfrac{25}{100}=\dfrac{h_{1}}{h_{2}}$

$h_{2}=4h_{1}$

Hence proved.