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Standard XII

Physics

Multiplication with Vectors

A ball is thr...

Question

A ball is thrown downwards with a speed of $20 m / s$ from the top of a building $150 m$ high and simultaneously another ball is thrown vertically upwards with a speed of $30 m / s$ from the foot of the building. Find the time after which both the balls will meet. $(g = 10 m / s^{2})$

4 s

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5 s

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3 s

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1 s

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Solution

The correct option is C $3 s$
The distance covered is given as
$s_{1} = 20 t + 5 t^{2}$
$s_{2} = 30 t - 5 t^{2}$

$s_{1} + s_{2} = 150 \Rightarrow 150 = 50 t \Rightarrow t = 3 s$
Hence, the time after which both the balls will meet is 3 s.

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A ball is thrown downwards with a speed of 20m/s from the top of a building 150m high and simultaneously another ball is thrown vertically upwards with a speed of 30m/s from the foot of the building. Find the time after which both the balls will meet. (g =10 m/s2)

The correct option is C 3sThe distance covered is given ass1=20t+5t2s2=30t−5t2s1+s2=150⇒150=50t⇒t=3sHence, the time after which both the balls will meet is 3 s.

A ball is thrown downwards with a speed of $20 m / s$ from the top of a building $150 m$ high and simultaneously another ball is thrown vertically upwards with a speed of $30 m / s$ from the foot of the building. Find the time after which both the balls will meet. $(g = 10 m / s^{2})$

The correct option is C $3 s$
The distance covered is given as
$s_{1} = 20 t + 5 t^{2}$
$s_{2} = 30 t - 5 t^{2}$

$s_{1} + s_{2} = 150 \Rightarrow 150 = 50 t \Rightarrow t = 3 s$
Hence, the time after which both the balls will meet is 3 s.