Question

Let h = (v0^2)/4.9 sin(theta)cos(theta) model the horizontal distance in meters traveled by a projectile? If the initial velocity is 52 meters/second, which equation would you use to find the angle needed to travel 200 meters?

A) 275.92sin(2Theta)=200
B) 551.84sin(2Theta)=200
C) 200sin(2 Theta)=200
D)10.61sin(2THeta)=100

Answer 1

A) 275.92sin(2Theta)=200

a p e x

Answer 2

The horizontal distance traveled by the projectile is given as

`d= (V_(0)^(2))/(4.9) sin\theta cos\theta`
where
V₀ = initial velocity, m/s

Note that
sin(2θ) = 2sinθ cosθ

Therefore, the horizontal distance traveled is

`d= (V_(0)^(2))/(4.9) ( (sin(2\theta))/(2) )= (V_(0)^(2))/(9.8) sin(2\theta)`

Because V₀ = 52 m/s, in order to travel a horizontal distance of 200 m, the correct equation to determine θ is
(52²/9.8) sin(2θ) = 200
or
275.92 sin(2θ) = 200

Answer: A