Question

From the top of a tower $100m$ in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of $25m{s^{ - 1}}$ . Find when and where the two balls will meet. Take $g = 9.5m{s^{ - 2}}$ .

Answer 1

Hint
Whenever an object is dropped from a height, its initial velocity is zero. The acceleration of the body is constant and is equal to the acceleration due to gravity. Displacement of an object moving with constant acceleration is given by $s = ut + \dfrac{1}{2}a{t^2}$ where $u$ is the initial velocity of the object, $a$ is its constant acceleration and $t$ is the time.

Complete step by step answer
As given in the question that one ball is dropped from a height, so whenever an object is dropped from a height, its initial velocity is zero. So, ${u_1} = 0$ . And another ball is projected vertically upwards with a velocity of $25m{s^{ - 1}}$ at the same time. So, ${u_2} = 25m{s^{ - 1}}$ as we take upward direction to be positive.
Both balls move with a constant acceleration equal to the acceleration due to gravity.
So, ${a_1} = {a_2} = - 9.5m{s^{ - 2}}$ . The negative is due to the fact that the acceleration due to gravity always acts in downward direction.
Now, let the ball which has been thrown upwards meets the other ball at distance $x{\text{ m}}$ from the ground. So, the other ball will cover a distance of $\left( {100 - x} \right){\text{ m}}$.
As we know that displacement of an object moving with constant acceleration is given by
$s = ut + \dfrac{1}{2}a{t^2}$
Where, $u$ is the initial velocity of the object, $a$ is its constant acceleration and $t$ is the time. Let the balls meet in time $t$ sec.
So, for the ball which was dropped,
$ - \left( {100 - x} \right) = - \dfrac{1}{2}g{t^2}$ ……(i)
Now, for the ball which was projected upwards,
$x = 25t - \dfrac{1}{2}g{t^2}$ ……(ii)
Now, substituting the value of $x$ from equation (i) in equation (ii) we have
$100 - \dfrac{1}{2}g{t^2} = 25t - \dfrac{1}{2}g{t^2}$
On simplifying we have
$25t = 100$
So, $t = 4$
Now, putting this value of time in equation (ii) we have
$x = 25 \times 4 - \dfrac{1}{2} \times 9.5 \times {4^2}$
On simplifying we have
$x = 24$
Hence, the balls will meet at a distance $24{\text{ m}}$ from the ground in $4{\text{ second}}$ .

Note
When a body drops from a height its initial velocity is zero so it moves downward only because of the acceleration due to gravity. The body moves due to the only force acting on it i.e. gravitational force. This type of motion is known as Free Fall motion.