Projectile and velocity

Question

The range of a projectile fired at an angle theta with the horizontal and with an initial velocity of v0 feet per second is r = 1/32v0^2 sin 2theta, where r is measured in feet. Determine the expression for the range in terms of theta. 

Answer 1

vx = v0*cosθ

 

x(t) = vx*t = v0*cosθ*t

 

h(t) = h0 + v0*sinθ*t - gt^2 / 2

 

Range r is reached when y(t) = 0

 

Let h0 = 0, i.e. start from the ground

 

h(t) = v0*sinθ*t - gt^2 / 2 = t(v0*sinθ - gt/2) 

 

h(t) = 0 ⇒ v0*sinθ - gt/2 = 0  ⇒ t = 2v0*sinθ / g

 

r = x(t) = v0*cosθ*(2v0*sinθ / g) = 2(v0^2)*cosθ*sinθ / g

 

Trigonometric identity:  sin(2θ) = 2sinθcosθ ⇒ sinθcosθ =  sin(2θ) / 2

 

r = 2(v0^2)*sin2θ / 2g = v0^2 * sin2θ / g 

 

r(θ) =  v0^2 * sin2θ / g