Question

18. A ball is projected under gravity with velocity 20 m/s
at an angle with the horizontal such that horizontal
range is equal to the maximum height of projectile.
The range of projectile is (g = 10 ms-2)
​

Answer 1

answer : 18.82 m (approximately)

explanation : Let the ball is projected at an angle θ with horizontal.

a/c to question,

horizontal range = maximum height

⇒u²sin2θ/g = u²sin²θ/2g

⇒2sinθcosθ = sin²θ/2

⇒4 = sinθ/cosθ

⇒tanθ = 4

so, sinθ = 4/√17 and cosθ = 1/√17

now, range of projectile = u²sin2θ/g

= u²(2sinθcosθ)/g

= (20)²(2 × 4/√17 × 1/√17)/10

= (400)(8/17)/10

= (320)/17

≈ 18.82 m

Answer 2

$\huge\bold\purple{Answer:-}$

18.82 m