Question

From the top of a tower of 100m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground so that it reaches just the top of the tower,at what height do the balls pass each other?

Answer 1

$\huge\text{\underline{Answer}}$

21.6 m above the ground.

$\sf{\underline{Explanation }}$

Let the initial velocity of ball be U since it is thrown from upward I. e. U=0.

Height of tower = 100 m

Let the ball meet at a height of x m

And Time taken be T

Then by using Equation of motion

$\bold{ h = ut + \frac{1}{2} g {t}^{2} }$

here ut = 0

$\implies$ $\bold{ x = \frac{1}{2} \times 9.8 \times {t}^{2} }$

$\implies$ $\bold{ x = 4.9 {t}^{2} }$

Now, Second case

When the ball is projected vertically upward it's final velocity V = 0.

And height be (100 - x ) m

and initial velocity is 25 m/s

$\implies$ $\bold{(100 - x) = ut - \frac{1}{2} g {t}^{2} }$

$\implies$ $\bold{(100 - x) = 25t - \frac{1}{2} \times 9.8 \times {t}^{2}}$

$\implies$ $\bold{(100 - x) = 25t - 4.9 \times {t}^{2} }$

Putting the value of x in equation

$\implies$ $\bold{100 - 4.9 {t}^{2} = 25t - 4.9 {t}^{2} }$

$\implies$ $\bold{25 t = 100}$

$\implies$ $\bold{t = 4 sec. }$

Now putting the value of t in equation. 1

$\implies$ $\bold{x = 4.9× 16}$

$\implies$ $\bold{x = 78.4m}$

The height when ball meet is $\bold{100 - 78.4}$

$\implies$ $\bold{21.6 m }$

Answer 2

Answer:

Explanation:

75m from ground