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Question

The height of the tower is $$200m$$ a ball thrown up with a velocity of $$50m/s$$ , another ball dropped from a top of a tower, when and where they meet?


Solution

The distance travelled by the body dropped from the height is given as

$$s = ut + \dfrac{1}{2}g{t^2}$$

$$s = 50t - 5{t^2}$$

Both the body will together travel the distance of $$200\;m$$

Hence

$$200 = \left( {50t - 5{t^2}} \right) + 5{t^2}$$

Time taken is$$4\;\sec $$

Distance travelled by the falling stone is

$$s = \dfrac{1}{2}g{t^2}$$

$$s = 5 \times 8 = 80\;m$$

The distance travelled by the projected body is

$$s = 50 \times 4 - 5 \times {4^2}$$

$$s = 120\;m$$

So both body will meet at $$120\;m$$ above the ground.

 


Physics

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