Question

VIII PHISICS
A steel ball of mass 50g is thrown vertical
downward with a velocity of 15m/s from a heich
of 20m. It buries itself in the sandy ground to a
depth of 20cm. The magnitude of average force
exerted on the ball by the sand is nearly (10m/s^2)
1) 78.1N
2) 78.6N
3) 77.6N
4) 8.3N​

Answer 1

Answer:

77.6N (3) is answer of this quetion

Answer 2

                      → S o l u t i o n ←

Given :

• Mass of ball (m) = 50 g
• It' initial velocity (u) = 15 m/s
• Given height = 20m
• It buries itself in the sandy

        ground to a depth of =20 cm.

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To find :

• The Force

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How to solve ?

1. We have to find final velocity.
2. And next we have to find 'a'
3. After that we have to find 'F'

======================

Let's sove

1. We have to find final velocity.

To find final velocity we have to

use third equation of motion

v² = u² + 2as

v² = (15)² + 2 × 10 × 20

v² = 225 + 20 × 20

v² = 225 + 400

v² = 625

v = √625

∴ v = 25 m/s

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Now we have to find that when the ball comes at rest

in sany ground after travelling the distance of

20 cm

Here, we again use third equation of motion

v² = u² + 2as

0² = 25² - 2 × a × 0.20

0 = 625 - 0.4a

a = 625/0.4

∴ a = 1562.5  m/s²

======================

Now,

The force exerted by the sand is F

and the force exerted by the ball

is mg

∴ $\implies d=\dfrac{F-mg}{m}\\\\\implies1562.5=\dfrac{F-0.5}{0.05}\\\\\implies1562.5\times0.05=F-0.5\\\\\implies78.125=F-0.5\\\\\therefore F=78.125+0.5\\\\=\boxed{78.625\:N}$

======================

1) 78.1N

2) 78.6N ✔

3) 77.6N

4) 8.3N​