Question

A stone is projected obliquely in air with a velocity v. It is found that its horizontal range (R) is n times
the maximum height (H) reached by the stone. What is the angle of projection?
COS​

Answer 1

Explanation:

stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H

1

. When it is projected with velocity u at an angle (

2

π

−θ) with the horizontal, it reaches maximum height H

2

. The relation between the horizontal range R of the projectile, H

1

and H

2

is

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VIDEO EXPLANATION

ANSWER

We know that, H

1

=

2g

u

2

sin

2

θ

and H

2

=

2g

u

2

sin

2

(90−θ)

⇒H

2

=

2g

u

2

cos

2

θ

∵H

1

H

2

=

2g

u

2

sin

2

θ

×

2g

u

2

cos

2

θ

=

4

4

×

2g

U

2

sin

2

θ

×

2g

U

2

cos

2

θ

=

16g

2

(U

2

sin2θ)

2

[∵sin

2

θ=2sinθcosθ]

=

16

R

2

R

2

=6H

1

H

2

∴R=4

H

1

H

2

Answered By

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