Question

The speed of a projectile v reduces to 50%, on reaching the highest point above the ground. Then its range on the horizontal plane is

Answer 1

Answer:

Explanation:

At maximum height

velocity = uCosθ

u/2 = uCosθ

θ = 60°

R = u² Sin(2θ) / g

= u² Sin(2 × 60°) / g

= u²√3 / (2g)

Range on horizontal plane is u²√3 / (2g)

Answer 2

Answer:

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