Question

A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered a distance of 20 m?

Answer 1

If it is simply dropped then its initial velocity = 0 m/s.
$v^{2} - u^{2}$ = 2as,
v = ? , u = 0, a =g =acceleration due to gravity, s = distance traveled  = 20m
$v^{2}$ = 2*9.8*20
v = $\sqrt{392}$ = 19.8 m/s

Answer 2

Answer:

v= 20m/s

Explanation:

We have ,

u=0m/s

v=?

s= 20m

g=10m/s

${v}^{2} = u {}^{2} + 2as$

=

$v {}^{2} = 0 {}^{2} + 2 \times 10 \times 20 \\ v {}^{2} = 400 \\ v = 20ms$

Hope It you