Question

A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 $m{{s}^{-2}}$, with what velocity will it strike the ground and time after which it will strike the ground?
$\text{A}\text{. }20m{{s}^{-1}},2s$
$\text{B}\text{. }24m{{s}^{-1}},\sqrt{2}s$
$\text{C}\text{. 1}0m{{s}^{-1}},4s$
$\text{D.}$ None of the above

Answer 1

Hint: Analyse the given problem. Write down the initial and final velocities of the ball before hitting the ground. Also the acceleration and displacement of the ball. Then figure which of the kinematic equations will help us to solve the problem.

Formula used:
$2as={{v}^{2}}-{{u}^{2}}$
$v=u+at$
$W=\Delta K.E$

Complete step by step answer:

It is given that the ball is dropped from a height of 20m. The acceleration of the body is given to be 10 $m{{s}^{-2}}$. We are supposed to find the time after which the ball will strike the ground from its release point and the velocity of the ball with which it will strike the ground.

For this we can use the kinematic equations :-
(i) $2as={{v}^{2}}-{{u}^{2}}$,
where a is the acceleration of the body, s is the displacement of the body, u is the initial velocity of the body (i.e. at time t=0) and v is the final velocity of the body (i.e. the velocity with which the ball will hit the ground).
(ii) $v=u+at$ ,
where u, v and a are the initial velocity, final velocity and acceleration of the given body. t is the time taken for the body to attain the final velocity from its initial velocity.
With these two equations, we will be able to solve the given problem.
First let us find the final velocity v (velocity with which the ball will strike the ground) of the ball with equation (i) and then substitute the value of v in equation (ii) to find the time t after which the ball will strike the ground.
So, let's begin by substituting the values of u, a and s in equation (i).
Consider the upward vertical direction as positive and downwards vertical direction as negative.
Therefore, u=0 (since the body is dropped from rest ).
$a=-10m{{s}^{-2}}$
$s=-20m$
(Since the motion of the ball is in downward direction and the velocity is uniformly increasing downwards, the displacement (s) and the acceleration (a) of the ball will be negative).
Now, substitute the values of u, a and s in equation (i).
$\Rightarrow 2(-10)(-20)={{v}^{2}}-{{(0)}^{2}}$.
$\Rightarrow 2\times 10\times 20={{v}^{2}}$
$\Rightarrow {{v}^{2}}=2\times 10\times 20=400\Rightarrow v=\sqrt{400}=\pm 20m{{s}^{-1}}$.
Since, we know that the final velocity of the ball will be in downward direction, v will be negative value.
Therefore, $v=+20m{{s}^{-1}}$ is discarded and we consider $v=-20m{{s}^{-1}}$.
Substitute the value of $v=-20m{{s}^{-1}}$ in equation (ii).
Therefore we get,
$-20=0+(-10)t$
$\Rightarrow 20=10t\Rightarrow t=2s$
Therefore, the velocity of the ball with which it strikes the ground is $20m{{s}^{-1}}$ downwards and the time taken for the ball to hit the ground is 2 seconds..
Hence, the correct option is A.

Note: We know that when a body falls or is dropped from a height, it experiences a gravitational pull towards earth. Due to this force, the body accelerates with an acceleration of g, which is acceleration due to gravity. The value of g is almost equal to 10$m{{s}^{-2}}$. Since the acceleration of the body is given to be 10$m{{s}^{-2}}$downwards , we do not have to consider the acceleration due to gravity.
The velocity with which the ball hits the ground can be found by an alternative method. We can use the work-energy theorem, which states that work done on a body is equal to the change its kinetic energy i.e. $W=\Delta K.E$.
Here, the change in kinetic energy will be $\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}$.
The work done on a body is the product of the force and its displacement parallel to the force i.e. W=Fs. we know that F=m. Therefore, W=mas.
This implies that $\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}=mas$.
$\Rightarrow \left( {{v}^{2}}-{{u}^{2}} \right)=2as$
From here, it is the same procedure that we have done in the step-by-step answer section.