Need help on this question Id really appreciate it if you gu

Need help on this question. I\'d really appreciate it if you guys show me how to solve this problem step by step.

The speed of a projectile when it reaches its maximum height is 0.42 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Solution

for a projectile

vo = initial velocity

horizantal component = vox = vo*costheta

vertical componenet = voy = vo*sintheta

ax = 0

ay = -g

along vertical direction

let H be the maximum height

from equations of motion

vy^2 - voy^2 = 2*g*H

vy = 0 at max heigth

H = voy^2/2g

at half of the maximum heighti.e y = H/2 = voy^2/4g, v1 be the speed

from equation of motion

v1y^2 - voy^2 = 2*g*voy^2/4g

v1y^2 = 3*voy^2/2

v1x = vox

v1^2 = v1x^2 + v1y^2

v1^2 = vox^2 + 1.5voy^2 .....(1)

at the maximum height the vertical component of velocity becomes zero .it has only horizantal componenet vox = vo*costheta

speed at maximum height = v = vo*costheta

given

v = 0.42*v1

v^2 = 0.42^2*v1^2

vox^2 = 0.42^2*(vox^2 + 1.5voy^2)

vox^2 (1-0.1764) = 0.2646*voy^2

voy/vox = 1.764

vo*sintheta/vo*costheta = 1.764

tan theta = 1.764

tehta = 60.45 degrees <----answer