since the range is
R = (v^2 sin2θ)/g, I get
60^2 * sin(1.094)/16 = 200.0
Looks like you lost a factor of 2 there, since sin2θ = 2sinθcosθ
Recall that the equation for the horizontal distance h in feet of a projectile with initial velocity v0 and initial angle theta is given by h=v0^2/16 sin theta cos theta .
a. Assume the initial velocity is 60 ft/second. What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet?
Here is my answer. Can anyone check if it's correct. Thank you.
60^2/16 sinᶿ cosᶿ=100
yields sin^2 ᶿ=.8889
theta=.547 radians
theta= 33.16 degrees
2 answers
furry sack of balls