Question

the equation for the horizontal distnace h in feet of a projectile with initial velocity v(subscript o) and initial angle θ is given by h=v(subscrit 0)^2/16sinθcosθ

Assume the initial velocity is 60 feet/seconds. What initial angle will you need to ensure that the horizontal distnace will be exactly 100 feet?

Answer 1

\[v_{o} = 60, h = 100\]
\[100 = \frac{60^{2}}{16\sin(\theta)\cos(\theta)}\]
Also use trig identity
\[\sin(2\theta) = 2\sin(\theta)\cos(\theta)\]
Substitute this in
\[\rightarrow 100 = \frac{60^{2}}{8\sin(2\theta)}\]
\[800\sin(2\theta) = 60^{2}\]
\[\sin(2\theta) = \frac{60^{2}}{800} = 4.5\]
hmm that cant be right, range of sin is -1 to 1
did you copy the equation correctly ??

Answer 2

maybe its
\[h = \frac{v_{o}\sin(\theta)\cos(\theta)}{16}\]
??

Answer 3

oh well, i'll assume its
\[h = \frac{v_{o}^{2}\sin(\theta)\cos(\theta)}{16}\]
Follow the same steps as above
\[\rightarrow \sin(2\theta) = 0.888\]
\[2\theta = 62.73\]
\[\theta = 31.365\]