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

Answer:

394.26 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v=u+at\\\Rightarrow 0=44-9.8\times t\\\Rightarrow \frac{-44}{-9.8}=t\\\Rightarrow t=4.49 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=44\times 4.49+\frac{1}{2}\times -9.8\times 4.49^2\\\Rightarrow s=98.77\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 98.77=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{98.77\times 2}{9.8}}\\\Rightarrow t=4.49\ s$

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

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{h\times 2}{9.8}}$

The times are equal so,

$8.97=\sqrt{\frac{h\times 2}{9.8}}\\\Rightarrow h=\frac{8.97^2\times 9.8}{2}\\\Rightarrow h=394.26\ m$

The height is 394.26 m