Question

A ball is gently dropped from a height of 20m . If its velocity increases uniformly at rate of 10m/s-2 , with what velocity will it strick the ground? After what time will it strike the ground?

Answer 1

Given :

▪ Initial velocity = zero

▪ Height = 20m

▪ Acceleration due to gravity = 10m/s²

To Find :

▪ Final velocity of ball.

▪ Time taken by ball to reach at the ground.

Concept :

→ Since, acceleration due to gravity has said to be constant through the journey, we can easily apply equations of motion to solve this type of questions.

→ For a body falling freely under the action of gravity, g is taken positive.

Calculation :

⚾ Final velocity :

$\implies\bf\:v^2-u^2=2gH\\ \\ \implies\sf\:v^2-(0)^2=2(10)(20)\\ \\ \implies\sf\:v^2=400\\ \\ \implies\underline{\boxed{\bf{\blue{v=20mps}}}}\:\gray{\bigstar}$

⚾ Time taken by ball :

$\dashrightarrow\bf\:H=ut+\dfrac{1}{2}gt^2\\ \\ \dashrightarrow\sf\:20=(0)t+\dfrac{1}{2}(10)t^2\\ \\ \dashrightarrow\sf\:t=\sqrt{\dfrac{2\times 20}{10}}\\ \\ \dashrightarrow\sf\:t=\sqrt{4}\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{t=2s}}}}\:\gray{\bigstar}$

Answer 2

Explanation:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms

−2

Time of fall, t=?

We know

v 2 −u 2 =2as

or v 2

−0=2×10×20=400 or v=20ms

−1

Now using v=u+at we have

20=0+10×t or t=2s