Question

A stone is dropped from top of a tower 300 m high and at the same time another is projected vertically upward with a velocity of 100ms−1. Find when and where the two stones meet?

Answer 1

Answer:

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Explanation:

height covered by the projectile from the ground

h=ut−21gt2

 h=100t−21×9.8×t2

 h=100t–4.9t2…(1)

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

300–h=21gt2

 300–(100t–4.9t2)=4.9t2…byequation(1)

 300=100t,t=3s

Thus particles meet each other at 3s

Height at which particles meet

h=100t–4.9t2

 h=100×3–4.9×9

h=255.9m