Question

A ball is dropped gently from a height of 20m. 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:

v= 20m/s

t= 2 sec

Explanation:

Initial Velocity of the ball (u) = 0

Distance or height of fall (s) = 20 m

Downward acceleration (a) = 10 m/s²

As we know,

2as = v² – u²

v² = 2as +  u²

v² = (2 x 10 x 20 ) + 0

v² =  400

Final velocity of ball (v) = 20 m/s

t = (v – u)/a

Time taken by the ball to strike (t) = (20 – 0)/10

t = 20/10

t = 2 seconds

Hope it helped you.

Answer 2

Explanation:

★ Concept :-

Here the concept of Equations of Motion has been used. As we see, that we are given the Distance travelled and the Acceleration of the ball. Then firstly, using the third equation of motion we will find out the final velocity of the ball to strike the ground. After that, using first equation of motion we will find out the time taken by the ball to strike the ground.

Let's do it !!!

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★ Equations Used :-

$\star\;\underline{\boxed{\bf{\pink{v^2\ =\ u^2\ +\ 2as.}}}}$

$\star\;\underline{\boxed{\bf{\pink{v\ =\ u\ +\ at.}}}}$

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★ Solution :-

Given,

➷ Initial velocity, u = $\sf {0ms^{-1}.}$

➷ Distance travelled, s = $\sf {20m.}$

➷ Acceleration, a = $\sf {10ms^{-2}.}$

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~ Case 1 : From the third equation of motion ::

That is,

✪ v² = u² + 2as ✪

⦾ By applying the values, we get :-

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\blue{v^2\ =\ u^2\ +\ 2as}}$

$\sf \ {\blue{{or,}\: \: \: \: \: \: \: \: \: {v^2\ =\ (0ms^{-1})^2\ +\ 2 \times 10ms^{-2} \times 20m}}}$

$\sf \ {\blue{{or,}\: \: \: \: \: \: \: \: \: {v^2\ =\ 400m^2 s^{-2}}}}$

$\sf \ {\blue{{or,}\: \: \: \: \: \: \: \: \: {v\ =\ 20ms^{-1}.}}}$

∴ Hence, the ball will strike the ground with velocity of $\sf {20ms^{-1}.}$

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~ Case 2 : From the first equation of motion ::

That is,

✪ v = u + at ✪

⦾ By applying the values, we get :-

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\orange{v\ =\ u\ +\ at}}$

$\sf \ {\orange{{or,}\: \: \: \: \: \: \: \: \: {20ms^{-1}\ =\ 0ms^{-1}\ +\ 10ms^{-2} \times t}}}$

$\sf \ {\orange{{or,}\: \: \: \: \: \: \: \: \: {t\ =\ \dfrac{20ms^{-1}}{10ms^{-2}}}}}$

$\sf \ {\orange{{or,}\: \: \: \: \: \: \: \: \: {t\ =\ 2s.}}}$

∴ Hence, the time taken by the ball to strike the ground is 2 seconds.

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★ More to know :-

➥ Motion :- Change in position of an object with respect to a fixed position with time is called motion.

➥ Rest :- When the position of a body with respect to its surroundings does not change with time is called rest.

➥ Velocity :- The speed of an object moving in a definite direction is called velocity.

➥ Acceleration :- Rate of change of velocity with time is called acceleration.

➥ Distance :- It is the actual length of the path travelled by a body.