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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