Question

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

Answer 1

☞ Given,

• initial velocity of ball, u=0

• Final velocity of ball, v=?

Distance through which the balls falls, s=20m

☞ Acceleration

$a=10 \: m/s </p><p>−2$

• Time of fall, t=?

☞ We know

$v {}^{2} - u {}^{2} = 2as$

$v </p><p> {}^{2} </p><p> −0=2×10×20=400$

$v=20ms$

☞ Now using v=u+at we have

$20=0+10×t$

$t=2s$

Answer 2

Diagram :-

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Solution :-

Given ,

• Initial velocity = 0 m/s
• Height = 20m
• Speed = 10 m/s

We need to find ,

• Velocity = ?
• Time taken to strike the ground = ?

Note :- Here initial velocity of the ball will be zero because the ball was dropped from rest , why it is dropped fro. rest because the gently dropped from hand .

Finding Velocity using kinematic equation,

• v² - u² = 2gH

Substituting known values we have ,

⇒ v² - 0² = 2 ( 10 ) ( 20 )

⇒ v² - 0 = 20(20)

⇒ v² = 20²

⇒ v = $\sf\sqrt{20^2}$

⇒ velocity = 20 m/s

Now , finding time taken to strike ground using the kinematic equation .

• v = u + at

Substituting known values we have ,

⇒ 20 = 0 + 10 ( t )

⇒ 20 = 10t

⇒ t = $\sf\dfrac{20}{10}$

⇒ Time taken to strike the ground = 2s

Hence ,

• Velocity = 20m/s
• Time taken to strike the ground = 2s