the vertical component of the velocity is v sin 52°, so solve for v in
790 - v*sin52°*6 - 4.9*6^2 = 0
Then use v*cos52° for the horizontal speed.
A plane, diving with constant speed at an angle of 52.0° with the vertical, releases a projectile at an altitude of 790 m. The projectile hits the ground 6.00 s after release. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.)
(a) What is the speed of the aircraft?
(b) How far did the projectile travel horizontally during its flight?
(c) What were the horizontal and vertical components of its velocity just before striking the ground?
1 answer
1 answer