Question:
A projectile is thrown from ground with a velocity u at an angle theta with the horizontal.
Obtain the expression for maximum height and horizontal range of projectile.
Find the angle of projection at which horizontal range and maximum height of projectile are equal.
Answer:
When projectile is thrown from ground, then-
Range R= u2 sin2θ/g
Height H = u2 sin2θ/2g
When Range is equal to Height,
R=H
u2 sin2θ/g = u2 sin2θ/2g
sin2θ = sin2θ/2
2 sinθcosθ = sin2θ/2
4 cosθ = sinθ
4 = tanθ
So angle θ= tan-1(4) for which R= H
= 75.96º
Not what you are looking for? Go ahead and submit the question, we will get back to you.
