Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms-2 , with what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

Answer :

Here

Initial velocity (u) = $0ms^{-1}$

Distance travelled (s) = 20 m

Accelaration (a) = Rate of change of velocity = $10ms^{-2}$

Final velocity (v) = ?

Time taken (t) = ?

Case 1 : From the equation of motion

⇒  $v^2 = u^2 + 2as$

⇒ $v^2 = (0 ms^{-1} ) ^2 + 2* 10ms^{-2}* 20m$

⇒ $v^2 = 400m^2s^{-2}$

⇒  $v = 20 ms^{-1}$

Case 2 : We know that

⇒  $v = u + at$

⇒  $20ms^{-1} = 0ms^{-1} + 10ms^{-2} * t$

⇒ $t = 20ms^{-1} / 10ms^{-2}$

⇒ $t = 2 s$

ANS

So the ball will strike the ground with velocity $20ms^{-1}$ after two seconds

                                                                                                                                                                                                                             

Hope it helps you

Answer 2

Given :

★ Initial velocity of ball, u = 0 m/s

★ Distance by which ball falls, S = 20 m

★ Acceleration, a = 10 m/s²

To find :

★ Velocity with which ball strikes the ground.

★ time taken.

SoluTion:

We know that,

$\large{\boxed{\sf{v^{2} - u^{2} = 2aS}}}$

Putting the given values,

$\longrightarrow$ v² - (0)² = 2 × 10 × 20

$\longrightarrow$ v² = 400

$\longrightarrow$ v = ±√400

$\longrightarrow$ v = ±20

Rejecting the negative value, we get,

$\longrightarrow$ v = 20 m/s

$\rule{200}2$

We also know that,

$\large{\boxed{\sf{v = u + at}}}$

Putting the values,

$\longrightarrow$ 20 = 0 + 10t

$\longrightarrow$ 20 - 0 = 10t

$\longrightarrow$ 10t = 20

$\longrightarrow$ t = $\dfrac{20}{10}$

$\longrightarrow$ t = 2 s

Hence,

• Final Velocity = 20 m/s

• time taken = 2 s