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

The horizontal distance traveled by the projectile is given as
$d= \frac{V_{0}^{2}}{4.9} sin\theta cos\theta$
where
V₀ = initial velocity, m/s

Note that
sin(2θ) = 2sinθ cosθ

Therefore, the horizontal distance traveled is
$d= \frac{V_{0}^{2}}{4.9} ( \frac{sin(2\theta)}{2} )= \frac{V_{0}^{2}}{9.8} sin(2\theta)$

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

Answer 2

Answer:

A) 275.92sin(2Theta)=200

a p e x