Projectile Motion Max Height and Range

  • Thread starter gooby99
  • Start date
  • #1
gooby99

Homework Statement


A projectile is launched with initial speed v0 and angle θ over level ground. The projectile's maximum height H and horizontal range R are related by the equation R = 4H. Write an expression for the launch angle of the projectile.

Homework Equations


R=4H
Vxo=Vocos(θ)
Vyo=Vosin(θ)

The Attempt at a Solution


Knowing that maximum height H occurs at time .5t, so H = Vosin(θ) x .5t
Similarly horizontal range R occurs at time t, so R = Vocos(θ) x t
Multiplying both sides of H = Vyosin(θ) x .5t by 4 gives

4H = Vyosin(θ) x 2t

Because of the given R = 4H,

Vosin(θ) x 2t = Vocos(θ) x t.

All the Vo and t terms cancel leaving 2sin(θ) = cos(θ), or tan(θ) = 1/2.

Thus θ = arctan(1/2), but the answer given is arctan(1). I am off by a factor of 2 somewhere.

Thanks for your help!
 

Answers and Replies

  • #2
MarkFL
You've neglected the fact that gravity is acting on the projectile.

I would begin with the acceleration resolved into the ##x## and ##y## components.

##a_y=-g##

##a_x=0##

Integrating with respect to time ##t##, and using the initial velocity, what are your resolved velocities?
 

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