Question

A boy stands 19.6 m away from 9.8m height of cliff. If he throw the ball towards the cliff such that, it just passes above from it, find the velocity of projection of the ball

Answer 1

Answer:

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Answer 2

Answer:14m/s

Explanation:

we know that acceleration in x-axis is 0

thats why vx=ux=ucosα - (1)

velocity= vx =ux

R=2*19.6m (Because the ball will pass horizontally from the highest point of the cliff)

We also know the relation between R/H=4cotα

39.2/9.8=4cotα

cotα=1

α=45°

H=u²sin²α/2g=9.8m (Given)

u² sin²45°/2·10 = 98

u²=392

u=√392

u=14√2

velocity=ucosα=14√2·cos45° (from (1))

velocity=14m/s