Question

If a 15 kg ball takes 5 seconds to strike the ground when released from rest ,at what height was the ball dropped​

Answer 1

The height from which the ball is dropped is 125m.

Given:-

Mass of the ball = 15kg

Time taken to strike the ground = 5sec

Initial velocity = 0m/s

To Find:-

The height from which the ball is dropped.

Solution:-

We can easily calculate the value of the height from which the ball is dropped by using these simple steps.

As

Mass of the ball (m) = 15kg

Time taken to strike the ground (t) = 5sec

Initial velocity (u) = 0m/s

g = 10m/s²

From the given values, mass of body has no use, as we can find out the height by using the second equation of motion.

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

on putting all the values, we get

$h = 0 \times 5 + \frac{1}{2} \times 10 \times {5}^{2}$

$h = 0 + 5 \times 25$

$h = 5 \times 25$

$h = 125m$

Hence, The height from which the ball is dropped is 125m.

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Answer 2

The ball was thrown from a height of 125 metres.

Given,

Mass of the ball=15 kg

Time taken to strike the ground=5 seconds

Initial velocity=0(as it is released from rest).

To find,

from what height the ball was dropped.

Solution:

Let the height be h.

As the ball is thrown downwards so the displacement will be negative.

Displacement, s=-h

Initial velocity, u=0

Time= 5 s

Acceleration=-g(always)

$(g=10\frac{m}{s^{2} })$

According to the law of motion,

$s=ut+\frac{1}{2} at^{2}$

The height can be calculated as:

$-h=0X5+\frac{1}{2} (-10)5^{2} \\-h=-\frac{1}{2} X10X25\\h=25X5\\h=125 m.$

There is also a direct formula for the calculation of time in dropped(thrown) cases.

$t=\sqrt{\frac{2h}{g} }$.

We do not consider any air resistance, so the time is independent of the mass of the body.

But if it is to be considered then, the time is inversely proportional to the mass of the body.

The height from where the ball was thrown is 125 metre.

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