An object is thrown with projectile motion and its observed that the horizontal component of velocity in the maximum height is Vx= 15.0m/s. The projectile travels horizontally a distance x=3.00x10^2m. Determine the launch angle.
Vx is constant the whole time. There is no horizontal force so no horizontal acceleration
Vx t = range
15 t = 300
t = 20 seconds in the air
Now do the vertical problem
20 seconds in air so 10 seconds up
Vy = Vy initial - 9.81 t on earth
at top Vy is 0 so
Vy initial = 9.81 * 10 seconds
Vy initial = 98.1 m/s
so tan theta = 98.1 / 15
theta = 81.3 deg