Question

A ball is gently dropped from a height of 20m. If its velocity increase uniformly at the rate of 20ms_2 with what velocity will it strike the ground

Answer 1

Hey,

Initial velocity (u) = 0
Acceleration (a) = 20 m/s^2
Distance (Height s) = 20 m

Final velocity (v) = ?

Using third equation of motion, we can find the answer.

v^2 - u^2 = 2as
v^2 = 2× 20×20
v^2 = 800
v = √800
v=28.28 m/s

Hope it helps...!!!

Answer 2

Hi there !!
Here's your answer

Given,
Distance or height (s) = 20m
Acceleration (a) = 20m/s²
Initial velocity (u) = 0 [ since the ball will Start in motion after dropping from a point]

Final velocity (v) = ?
Since the time is not given, the 1st equation can't be used.
Therefore,
We will use the 3rd equation of motion which states that :

2as = v² - u²
2(20)(20) = v² - 0²
800 = v² - 0
800 + 0 = v²
v² = 800
v = ✓800
v = 28.28

Thus,
It will strike the ground with a velocity of 28.28m/s or the final velocity will be 28.28m/s

__________________________

Hope it helps :D