Question

if the greatest height of a projectile is 1/4 of the range.what is the angle of projectile

Answer 1

We know the expression for maximum height of a projectile.

H = u² sin²θ / 2g

And we know the expression for the horizontal range too.

R = u² sin(2θ) / g

Then, given that,

H = R / 4

u² sin²θ / 2g = u² sin(2θ) / 4g

sin²θ = sin(2θ) / 2

sin²θ = sinθ cosθ [sin(2θ) = 2 sinθ cosθ]

sin²θ - sinθ cosθ = 0

sinθ (sinθ - cosθ) = 0

Since 0° < θ < 90°, sinθ can't be 0. Thus,

sinθ = cosθ

=> θ = 45°.

Hence the angle of projectile is 45°.

#answerwithquality

#BAL

Answer 2

Answer:

In projectile thrown at angle θ Range R and maximum height H are given as :

Range, R=

g

u

2

(sin2θ)

=

g

u

2

2sinθcosθ

Maximum Height, H=

2g

u

2

sin

2

θ

Given: H=R

g

u

2

(2sinθcosθ)

=

2g

u

2

sin

2

θ

4cosθ=sinθ

tanθ=4

θ=76

0