Question

The speed of a projectile when it reaches its maximum height is 0.53 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Answer 1

let vox and voy are the componenst of the initial velocity.

let h is the maximum height reached

so, h = voy^2/(2*g)

voy^2 = 2*g*h

let vy is the speed of projectile at half maximum.

use, vy^2 - voy^2 = 2*(-g)*(h/2)

vy^2 - voy^2 = -2*g*h/2

vy^2 - 2*g*h = g*h

vy^2 = g*h

= voy^2/2

speed of of projectile at maximum height = 0.53*speed at half its maximum height

vox = 0.53*sqrt(vox^2 + vy^2)

vox = 0.53*sqrt(vox^2 + voy^2/2)

vox*sqrt(2) = 0.53*sqrt(2*vox^2 + voy^2)

2*vox^2 = 0.53^2*(2*vox^2 + voy^2)

2*vox^2 = 0.5618*vox^2 + 0.2809*voy^2

vox^2*(2 - 0.5618) = 0.2809*voy^2

voy/vox = sqrt((2-0.5618)/0.2809)

= 2.263

angle of projection, theta = tan^(voy/vox)

= tan^-1(2.263)

= 66.1 degrees <<<<<<<<-------------Answer