Question

A ball which is thrown vertically upwards reaches the
roof of a house 100 m high. At the moment this ball
is thrown vertically upward, another ball is dropped
from rest vertically downwards from the roof of the
house. At which height will the balls pass each other?
(g = 9.8 m/s2)​

Answer 1

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Let the point where both ball will meet be Z m.

Now we know that both balls were throwed at the same time.

And as value of (g) remains constant this means both traveled same time to meet.

Now, some Important keywords :

Let's say :-

• X = height traveled by dropped ball
• Y = height traveled by upper thrown ball.
• v = their velocity

Now we know that the ball is dropped from top of roof which is 100m in height , so the sum of height both ball will teach will be 100.

Therefore, X + Y = 100m

Now according to third equation of motion.

${v}^{2} - {u}^{2} = 2as$

Since ball was thrown from rest thus the initial velocity of ball is 0.

Here (v) is the velocity of both bodies.

Now after inputting the known values in this equation we get :

${v}^{2} = 2 \times 9.8 \times 100$

${v}^{2} = 1960$

$v = \sqrt{1960}$

Now, we know that the product of velocity and time is the height traveled by them.

= vt, this is equal to 100m

vt = 100m

Now the time taken by balls :

$t = \frac{100}{v}$

$t = \frac{100}{ \sqrt{1960} } s$

Now we also have the time!

Now let's talk about at which height they will meet each other.

According to second equation of motion :

$s = ut + \frac{1}{2} \times a {t}^{2}$

For the dropped ball :

$x = 0 + \frac{1}{2} \times 9.8 \times \frac{100}{ \sqrt{1960} } \times \frac{100}{ \sqrt{1960} }$

$= \frac{1}{2} \times 9.8 \times \frac{100 \times 100}{1960}$

$= \frac{1}{2} \times 9.8 \times \frac{10000}{1960}$

$= 25m$

Thus now X = 25 m

Since X + Y = 100

Y = 100-X

Y = 100-25

Y =75m

Thus the dropped ball will meet at 25m., and the upward thrown ball will reach its 75m to meet.

______________________________________

BY√\____________ SCIVIBHANSHU

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