Question

A ball is dropped from the top of a tower hundred metre in height in the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s. find when and where two balls meet
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Answer 1

$\bold{\underline{\underline{Assume\::}}}$

Let the -

• time (t) when both the balls meet.

$\bold{\underline{\underline{Solution\::}}}$

Distance travelled by ball falling from the tower when it travels time (t).

We know that..

s = ut + 1/2at²

We have,

u = 25 m/s and a = 10m/s²

Substitute the known values in above formula

⇒ s = 0(t) + 1/2 × 10 × t²

⇒ s = 5t²

Now,

Distance travelled by the ball when thrown up from ground in time (t)

We have..

u = 0 m/s and a = -10 m/s² (as thrown upward)

⇒ s = ut + 1/2 at²

⇒ s = 25(t) + 1/2 × (-10) × t²

⇒ s = 25t - 5t²

The gap between both the balls is 100 m.

⇒ (5t²) + (25t - 5t²) = 100

⇒ 5t² + 25t - 5t² = 100

⇒ 25t = 100

⇒ t = 100/25

⇒ t = 4

Both the balls meet after 4 sec.

Now,

• Speed of falling ball from the top of tower = 5t²

⇒ 5(4)²

⇒ 80 m

• Speed of projected ball = 25t - 5t²

⇒ 25(4) - 5(4)²

⇒ 100 - 80

⇒ 20 m

∴ After 4 sec both the ball meet at a distance of 80 m.

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time=4\:sec}}$

${\bold{\therefore Distance=80\:m}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a ball is dropped from the top of a tower hundred metre in height in the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s.

• We have to find when and where two balls meet.

$\underline \bold{Given : } \\ \bold{For \:dropped \: ball : } \\ \implies Initial \: velocity(u )= 0 \\ \\ \implies Acceleration(a) = -10 \: m/{s}^{2} \\ \\ \bold{For \: projected \: ball : } \\ \implies Initial \: velocity(u) = 25 \: m/s \\ \\ \underline \bold{To \: Find : } \\ \implies Time \: and \: Distance \: of \: met = ?$

• According to given question :

$\bold{For \: dropped \: ball : } \\ \implies s = ut + \frac{1}{2} g {t}^{2} \\ \\ \implies - x = 0 \times t + \frac{1}{2} ( - g) \times {t}^{2} - - - - - (1) \\ \\ \bold{For \: projected \: ball : } \\ \implies s = ut + \frac{1}{2} {gt}^{2} \\ \\ \implies 100 - x = 25 \times t - \frac{1}{2} ( - g) {t}^{2} - - - - - (2) \\ \\ \bold{ Putting \: value \: of \: x \: in \: (2)} \\ \implies 100 - \cancel{ \frac{1}{2} g {t}^{2}} = 25 \times t - \cancel{\frac{1}{2} g {t}^{2} } \\ \\ \implies 25 \times t = 100 \\ \\ \implies t = \frac{100}{25} \\ \\ \bold{ \implies t = 4 \: sec} \\ \\ \bold{Putting \: value \: of \: t \: in \: (1)} \\ \implies \cancel{ -} x = \frac{1}{2} ( \cancel{- }10) \times 4 \times 4 \\ \\ \implies x = 5 \times 16 \\ \\ \bold{ \implies x = 80 \: m} \\ \\ \bold{\therefore After \: 4 \: sec \: both \: ball \: met }\\ \: \: \: \: \: \bold { and \: at \: 80 \: m.}$