Question

a projectile attains a maximum height of 100 m and has a horizontal range of 400 m .find the angle at which it is projected and the velocity of projection​

Answer 1

Answer:

45 degree and velocity

Explanation:

R tan\thetaθ =4H

H = 100m

R= 400m

tanθ= 4H /R

θ= tan-1 (4H/ R)

θ= tan-1 (4×100/400)

θ=tan-1 (400/400)

θ=tan-1( 1)

θ=45 degree

and velocity of projection will be

R=v² sin 2Θ/g

v²= R× g/ sin 2Θ

v²= 400× 9.8/ sin2 × 45

v²= 3920/ sin 90

v²=3920/ 1

v²= 3920

sqrt v²= Sqrt(3920)

v= 62.6099ms-1