For which pair of launch angles will two identical projectiles have equal ranges?
A. 19.24°, 80.54°B. 16.42°, 74.58°C. 60.23°, 29.77°D. 89.53°, 01.47°E. 42.42°, 47.59°

Question

For which pair of launch angles will two identical projectiles have equal ranges?
A. 19.24°, 80.54°B. 16.42°, 74.58°C. 60.23°, 29.77°D. 89.53°, 01.47°E. 42.42°, 47.59°

anonymous · Answer

If we establish the kinematic equations of motion, we see that$$y(t) = v_ot \sin(\theta) - {1 \over 2} g t^2$$and$$x(t) = v_o t \cos(\theta)$$

Setting $y(t) = 0$ yields $0=v_ot \sin(\theta) - {1 \over 2} g t^2$. If we solve for $t$, we obtain, by factoring, $$t = {2 v \sin(\theta) \over g}$$. Substitute this into our equation for $x(t)$. This yields$$x(t) = {2 v^2 \cos(\theta) \sin(\theta) \over g}$$Recall that $$\sin(2\theta) = 2\sin(\theta) \cos(\theta)$$Therefore, $$x = {v^2 \sin(2\theta) \over g}$$Assuming the same launch velocity and knowing that gravity is constant, we can establish the two angles from $\sin(2 \theta)$

anonymous · Answer

I'm am so confused... So I multiply sin 20 by the angles?

calculusfunctions · Answer

He's right. In simple terms multiply each value in each choice by two, and then find their respective sine ratios. See if they're equal.

anonymous · Answer

Okay thanks :)

anonymous · Answer

still dont understand uggh
XO