Question

A stone is thrown straight up from the ground with an initial speed of 44 m/s.At the same instant, a stone is dropped from a height of hmeters above ground level. The two stones strike the ground simultaneously. Find the heighth. The acceleration of gravity is 9.8 m/s2.Answer in units of m.

Answer 1

394.26 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

`v=u+at\\\Rightarrow 0=44-9.8* t\\\Rightarrow (-44)/(-9.8)=t\\\Rightarrow t=4.49 \s`

`s=ut+(1)/(2)at^2\\\Rightarrow s=44* 4.49+(1)/(2)* -9.8* 4.49^2\\\Rightarrow s=98.77\ m`

`s=ut+(1)/(2)at^2\\\Rightarrow 98.77=0t+(1)/(2)* 9.8* t^2\\\Rightarrow t=\sqrt{(98.77* 2)/(9.8)}\\\Rightarrow t=4.49\ s`

Total time taken by the stone to reach the ground is 4.49+4.49 = 8.97 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow h=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(h* 2)/(9.8)}`

The times are equal so,

`8.97=\sqrt{(h* 2)/(9.8)}\\\Rightarrow h=(8.97^2* 9.8)/(2)\\\Rightarrow h=394.26\ m`

The height is 394.26 m