Question

(in) A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate
of 10 m/s", after what time will it strike ground?
(a) 0.1s
(6) 1.0 s
(c) 0.2 s
(d) 2.0 s​

Answer 1

Given :-

Initial velocity of ball = 0 m/s

Height of the ball dropped = 20 m

Downward acceleration of the ball = 10 m/s

To Find :-

Time taken by the ball to strike ground.

Analysis :-

Here we are given with the initial velocity, distance and acceleration.

Using the third equation of motion, find the final velocity by substituting the values from the question.

Then using the first equation of motion find the time taken accordingly.

Solution :-

We know that,

• u = Initial velocity
• t = Time
• a = Acceleration
• s = Displacement
• v = Final velocity

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=2as=v^2-u^2}}$

Given that,

Acceleration (a) = 10 m/s

Displacement (s) = 20 m

Initial velocity (u) = 0 m/s

Substituting their values,

⇒ 2 × 10 × 20 = v² - 0²

⇒ 20 × 20 = v²

⇒ v² = 400

⇒ v = √400

⇒ v = 20 m/s

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Final velocity (v) = 20 m/s

Initial velocity (u) = 0 m/s

Acceleration (a) = 10 m/s

Substituting their values,

⇒ 20 = 0 + 10 × t

⇒ t = (v-u)/a

⇒ t = (20-0)/10

⇒ t = 20/10

⇒ t = 2 sec

Therefore, the ball will strike the ground at (d) 2.0 seconds.

Answer 2

Answer:-

$\small \bf \underline{Given:}$

★★ Initial velocity(u) = 0 (dropped gently)

★★ Acceleration due to gravity(a) = 10 m/s²

★★ Height(h) = 20 m

$\small \bf \underline{To \: find:}$

We need to find the time taken (t) by the ball to strike the ground

$\small \bf \underline{Solution:}$

Formula used:-

Second law of motion

Now let's solve the sum

$\bf h = ut + \frac{1}{2} a {t}^{2}$

$\bf→20 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2}$

$\bf→ t² = 4$

$\bf→ t = \sqrt{4}$

$\bf→ t = 2s$

Answer : Option (d) 2.0s is the correct answer

@BengaliBeauty

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