Question

When a ball is released from a top of building, the distance covered by it in 4 sec will be​

Answer 1

Given:-

• Time taken ,t = 4s
• Initial velocity ,u = 0m/s

To Find:-

• Distance covered by ball in 4 sec

Solution:-

According to the Question

As it is given that ball is released from top of building .

So, its initial velocity is zero .

Now, we know that the Earth will attract the ball some force called gravitational force and the acceleration produced in the ball will be 10m/s². You can also take the acceleration due to gravity 9.8m/s² if you want .

→ g = 10m/s²

Using Kinematics Equation

• S = ut + 1/2gt²

by substituting the value we obtain

→ S = 0×4 + 1/2× 10 × 4²

→ S = 0 + 5 × 16

→ S = 5 × 16

→ S = 80 m

• Hence, the distance covered by ball in 4 second will be 80 metres .

Answer 2

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Here Given that :

➸ u = 0

➪ u is initial velocity

➸ a = 10 m / s²

➪ a means Acceleration

➸ t = 4 second

➪ t means time

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To Find :-

• ➪ Distance travelled by ball in 4 second

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Explanation

• The gravitational force mean interaction between two objects with mass.

• Gravity pulls the ball toward the ground

• Since g=9.80m/s² on Earth but we take here

➪ g= 10 m/s²

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Note :

➭ S means distance here

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➭ Used this above equation and place the given value

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$\sf{➨ \: S = \frac{1}{2} \times 10 \times {t}^{2} }$

$\sf{➨ \: S = \frac{1}{ \cancel2} \times \cancel{ {10}}^{ \: \: 5}\times {t}^{2} }$

➤ Place the value of t that is

➤ t = 4 second

$\sf{➨ \: S =5 \times 4 \times 4}$

$\sf{➨ \: S =5 \times 16}$

$\bf{➨ \: S =80 \: m}$

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➯ Therefore

➲ 80 m is distance covered by ball in 4 second if ball is released from a top of building

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