Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s with what velocity will it strike the ground? After what time will strike to the ground

Answer 2 seconds ​

Answer 1

Solution :

• Initial Velocity (u) = 0

• Final Velocity (v) = ?

• Distance (s) = 20 m

• Acceleration (a) = 10 m/s²

Distance = Height

Now,

From 3rd equation of motion :

★ v² = u² + 2as

→ v² = (0)² + 2(10)(20)

→ v² = 0 + 400

→ v² = 400

→ v = √400

→ v = 20 m/s

Thus, the ball will strike the ground with a velocity of 20 metres per second. Let's find the time now :

From 1st equation of motion :

★ v = u + at

→ 20 = 0 + (10)(t)

→ 20 = 0 + 10t

→ 20 = 10t

→ t = $\sf\cancel\dfrac{20}{10}$

→ t = 2 s

$\therefore$ Ball will strike the ground after 2 seconds.

Answer 2

Correct Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s² with what velocity will it strike the ground? After what time will strike to the ground.

Solution

Given that, a ball is gently dropped from a height of 20 m. So, height is equal to distance covered by ball i.e. s = 20 m.

Also, its velocity increases uniformly at the rate of 10 m/s². (acceleration due to gravity is 10 m/s²).

We have to find the time taken by the ball.

Using the Third Equation Of Motion,

v² - u² = 2as

As ball is dropped from height, means the initial velocity of the ball is 0 m/s.

Substitute the known in the above formula,

→ v² - (0)² = 2(10)(20)

→ v² - 0 = 400

→ v² = 400

→ v = 20

Therefore, the final velocity of the ball is 20 m/s.

Now, using the First Equation Of Motion,

v = u + at

Substitute the values,

→ 20 = 0 + 10t

→ 20 = 10t

Divide by 10 on both sides,

→ 20/10 = 10t/10

→ 2 = t

Therefore, the time taken by the ball is 2 sec.