Question

SECTION -C 16. A ball is dropped from the top of a tower 100 m high and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/sec. Calculate where and when the two stones will will meet. (Take
g = 9.8 ms?) OR
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Answer 1

Given :-

A ball is dropped from the top of a tower 100 m high and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/sec. Calculate where and when the two stones will will meet

Solution :-

• Let us assume that the two stones will meet after a time "t"
• In the time "t" , the distance travelled by the first stone will be :-

        s₁ = $\sf\frac{1}{2}gt^{2}$   ----------------- (1)

• Now, the distance travelled by the ball thrown vertically upwards in the time t will be :-

       s₂ = ut $\sf\ - \frac{1}{2}gt^{2}$ = 25t $\sf\ - \frac{1}{2}gt^{2}$ ---------------- (2)

• Add equations (1) and (2)

      ⇒ s₁ + s₂ = $\sf\frac{1}{2}gt^{2} + 25t - \sf\frac{1}{2}gt^{2}$

    ⇒ s₁ + s₂ = 25t

• Here, s₁ + s₂ i.e distance travelled by first stone + second stone = height of the tower which is given as 100m

       ⇒ 100 = 25t

       ⇒ t = $\sf\frac{100}{25}$ = 4s

• Distance travelled by stone thrown vertically upwards = 25(4) - 4.9 × 16 = 21.6m

∴ The two stones meet after 4s at a height of 21.6m above the ground!

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Answer 2

Motion in a Straight Line

Uniformly Accelerated Motion

From the top of a tower 100