Question

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 10 metre per second . with what velocity will it strike the ground? after what time will it strike the ground?​

Answer 1

Solution:

Given:

➔ Initial velocity = 0 m/s

➔ Height = 20 m

➔ Acceleration = 10 m/s²

To Find:

➔ Final velocity

➔ Time

Formula used:

➔ v² = u² + 2as

➔ s = ut + ½at²

Now, firstly we will calculate final velocity by using 3rd equation of motion. So we get,

➔ v² = u² + 2as

➔ v² = 0 + 2 × 10 × 20

➔ v² = 400

➔ v = √400

➔ v = 20 m/s

Now, we will calculate time by using 2nd Equation of motion. So we get,

➔ s = ut + ½at²

➔ 20 = 0 + ½ × 10 × t²

➔ 40 = 10t²

➔ t² = 40/10

➔ t² = 4

➔ t = √4

➔ t = 2 sec.

Answer 2

Answer:

Explanation:

Given :-

Initial Speed of ball, u = 0 m/s

Height attain by ball, h = 20 m

Acceleration by ball, a = 10 m/s²

To Find :-

Final velocity

Time

Formula to be used :-

3rd Equation of Motion i.e v² = u² + 2as

2nd Equation of Motion i.e s = ut + ½at²

Solution :-

Putting all the value, we get

⇒ v² = u² + 2as

⇒ v² = 0 + 2 × 10 × 20

⇒ v² = 400

⇒ v = √400

⇒ v = 20 m/s

Now, we will find Time taken

⇒ s = ut + ½at²

⇒ 20 = 0 + ½ × 10 × t²

⇒ 40 = 10t²

⇒ t² = 40/10

⇒ t² = 4

⇒ t = √4

⇒ t = 2 sec.

Hence, the Final speed and time taken by ball is 20 m/s and 2 seconds.