Brandon launches a projectile at an angle of 75 above the horizontal, which strikes he ground a certain distance down range. fo what oher angle of launch at the same speed would the projectile land just as far away?
2 answers
d = V^2(sin(2µ))/g
Since the maximum range occurs at a launch angle of 45º, launch angles of +/- equal increments of µ on either side of 45º will result in the same horizontal distance traveled.
Therefore, the launch angle required to result in the same range of the projectile
becomes 45 - (75 - 30) = 15º.
Vsin(150) = Vsin30 = .5V
Therefore, the launch angle required to result in the same range of the projectile
becomes 45 - (75 - 30) = 15º.
Vsin(150) = Vsin30 = .5V