Question

A stone is dropped from a height of 19.6 metre calculate time taken to fall and how fast does it move at the end of this fall and acceleration after 1 second

Answer 1

Hey there! ☺☻☺

Given,
Initial velocity(u)= 0 m/s
Final velocity(v) = v m/s
Acceleration(a) = 9.8 m/s^2
Height(h) = 19.6 m

We know,
$v^2 - u^2 = 2gh\\v^2 = u^2 + 2gh\\v^2 = 0^2+2\times9.8\times19.6\\v^2 = 19.6 \times 19.6\\\boxed{\boxed{v = 19.6 \ m \ s^{-1}}}$

Now,For time. We apply:
$v = u+gt\\19.6 = 0 + 9.8\times t\\t = \frac{19.6}{9.8}\\\boxed{\boxed{t = 2}}$

Now, For acceleration after 1 sec: we apply:
$a = \frac{v-u}{t}\\\\a = \frac{19.9 - 0}{1}\\\\\boxed{\boxed{a = 19.6 \ m \ s^{-2}}}$

Hope It Helps You! ☺☻☺