Question

1 . A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 20 ms -¹ with what velocity will it strike the ground ? after what time will it ​

Answer 1

Given,

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 20 ms -²

To find :

With what velocity will it strike the ground and after what time will it ​ strike the ground?

Solution :

Given values,

Initial velocity, u = 0 m/s [As it falls freely]

Acceleration, a = 20 m/s²

Distance covered, s = 20 m

Now using 3rd equation,

⇒  v² = u² + 2as

⇒  v² = 0² + 2 * 20 * 20

⇒  v² = 800

⇒  v = √800

⇒  v = √(400 * 2)

⇒  v = √(20²) * 2

⇒  v = 20√2 m/s

∴ With a velocity of 20√2 m/s it will strike the ground.

Now using 1st equation of motion :

⇒  v = u + at

⇒  20√2 = 0 + 20t

⇒  28.28 = 20t

⇒  t = 28.28/20

⇒  t = 1.41 s

∴ After 1.414 s it will strike the ground.

Answer 2

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Have \; a \; glance \; at \; this \; before \; solving \; :-}}}$

Here the concept of Equations of Motions has been used. If we know the initial velocity (u), final velocity (v), acceleration (a), time (t), acceleration under gravity (g) and displacement (s), then we can apply the values and solve any question. Here firstly we will find the velocity with which ball striked the ground and then time taken by it.

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★ Formula Used :-

$\: \: \\ \qquad \large{\mapsto \: \: \boxed{\boxed{\bf{v^{2} \: - \: u^{2} \: = \: 2as}}}}$

This is the first equation of motion.

$\: \: \\ \qquad \large{\mapsto \: \: \boxed{\boxed{\bf{v \: - \: u \: = \: at}}}}$

This is the third equation of motion.

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★ Question :-

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 20 m/sec² with what velocity will it strike the ground ? After what time will it ?

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★ Solution :-

Given,

» Height from which the ball is dropped = h = 20 m

» Rate at which the velocity of ball increases = a = 20 m/s²

» Initial velocity of the ball = u = 0

(since the ball is dropped from an height, so its energy will be zero, and then its velocity starts increasing)

• Let the final velocity of the ball with which it strikes the ground be 'v' m/sec.

• Let the time taken by the ball to strike the ground be 't' second.

~ For final velocity of the ball :-

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: v^{2} \: - \: u^{2} \: = \: 2as}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: \bf{v^{2}} \: - \: 0^{2} \: = \: 2 \: \times \: 20 \: ms^{-2}\: \times \: 20 m}}$

➣ v² = 2 × 20 m/sec² × 20 m

➣ v² = 800 m²/sec²

$\: \: \\ \qquad \large{\sf{\bf{v} \: = \: \sqrt{800 \: m^{2}s^{-2}} \: = \: \underline{20\sqrt{2} \: \: ms^{-1}} \: \: or \: \: \underline{28.28 \: \: ms^{-1}}}}$

$\: \\ \large{\rm{\boxed{\boxed{\leadsto \: \: v \: = \: \bf{28.28 \: ms^{-1}}}}}}$

~ For time taken by the ball :-

$\: \: \\ \qquad \large{\bf{\Longrightarrow \: \: \: v \: - \: u \: = \: at}}$

$\: \: \\ \qquad \large{\sf{\Longrightarrow \: \: \: 28.28 \: ms^{-1}\: - \: 0 \: ms^{-1} \: = \: 20 \: ms^{-2} \: \times \: t}}$

➣ 28.28 = 20 × t

$\: \: \\ \qquad \large{\Longrightarrow \: \: \: \bf{t} \: = \: \dfrac{28.28 \: \cancel{ms^{-1}}}{20 \: \cancel{ms^{-2}}} \: = \: \underline{\bf{1.414 \: seconds}}}$

$\: \\ \large{\rm{\boxed{\boxed{\leadsto \: \: t \: = \: \bf{1.414 \: seconds}}}}}$

$\: \: \\ \large{\underline{\underline{\rm{\mapsto \: \: Thus, \: the \: final \: velocity \: of \: the \: ball \: is \: \boxed{\bf{28.28 \: ms^{-1}}} \: and \: the \: time \: taken \: by \: the \: ball \: is \: \boxed{\bf{1.414 \: sec}}}}}}$

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$\: \: \qquad \qquad \large{\tt{\underbrace{Aid \; to \; Memory} \; \; :-}}$

$\: \\ \mapsto \: \: \sf{s \: = \: ut \: + \: \dfrac{1}{2} \: at^{2}}$

This is second equation of motion.

• Some other formulas :-

$\: \\ \mapsto \: \: \sf{s_{n_{th}} \: = \: u \: + \: \dfrac{a}{2} \: (2n \: - \: 1)}$

$\: \\ \mapsto \: \: \sf{s \: = \: \dfrac{u \: + \: v}{2} \: \times \: t}$

$\: \\ \mapsto \: \: \sf{s \: = \: vt \: - \: \dfrac{1}{2} \: at^{2}}$

$\: \\ \mapsto \: \: \sf{v \: = \: \sqrt{2gh} }$

$\: \\ \mapsto \: \: \sf{h \: = \: \dfrac{u^{2}}{2g}}$

$\: \\ \mapsto \: \: \sf{t \: = \: \dfrac{u}{g}}$

$\: \\ \mapsto \: \: \sf{t_{total} \: = \: \dfrac{2u}{g}}$

$\: \\ \mapsto \: \: \sf{a \: = \: \dfrac{v \: - \: u}{t}}$

$\: \\ \mapsto \: \: \sf{a_{av} \: = \: \dfrac{v_{2} \: - \: v_{1}}{t_{2} \: - \: t_{1}} \: = \: \dfrac{\Delta v}{\Delta t}}$

$\: \\ \mapsto \: \: \sf{v_{av} \: = \: \dfrac{x_{2} \: - \: x_{1}}{t_{2} \: - \: t_{1}} \: = \: \dfrac{\Delta x}{\Delta t}}$

$\: \\ \mapsto \: \: \sf{v \: = \: \dfrac{s}{t}}$

$\: \\ \mapsto \: \: \sf{s \: = \: v \: \times \: t}$

$\: \\ \mapsto \: \: \sf{t \: = \: \dfrac{s}{v}}$

$\: \\ \mapsto \: \: \sf{\omega \: = \: \dfrac{\theta}{t} \: = \: \dfrac{\theta _{2} \: - \: \theta _{1}}{t_{2} \: - \: t_{1}}}$

$\: \\ \mapsto \: \: \sf{\omega \: = \: \dfrac{2 \pi}{T} \: = \: 2 \pi \nu}$

*Note :- Here the value 28.28 is equal to 20√2 only. Just if we remove square root, then we get √2 = 1.14 (approx). Then, we get 20√2 = 28.28

So both the answers are correct. You can use both the values either 20√2 m/sec or 28.28 m/sec as the answer.