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Question

A stone is dropped from a height of $$300$$m and at the same time another stone is projected vertically upwards with a velocity of $$100$$ $$ms^{-1}$$. Find when and where the two stones meet.


Solution

height covered by the projectile from the ground

$$h=ut-\dfrac { 1 }{ 2 } gt^2$$

 $$h=100t-\dfrac { 1 }{ 2 } \times 9.8 \times t^2$$

 $$h=100t–4.9t^2…(1)$$

Distance covered by the projectile thrown down in time t will be, since it falls down from the rest$$ u= 0$$

$$300–h=\dfrac { 1 }{ 2 } gt^2$$


 $$300–(100t–4.9t^2)=4.9t^2…by equation(1)$$


 $$300=100t, t=3s$$

Thus particles meet each other at $$3s$$
Height at which particles meet

$$h=100t–4.9t^2$$

 $$h=100 \times3–4.9 \times 9$$

$$h=255.9m$$







Physics

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