An airplane, diving at an angle of 50.0° with the vertical, releases a projectile at an altitude of 554. m. The projectile hits the ground 8.00 s after being released. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction.) (a) What is the speed of the aircraft?

Speed of the aircraft = 36.64 m/s

Explanation:

Consider the vertical motion of the projectile,

We have equation of motion s = ut+0.5at²

Let the velocity of plane be v.

Vertical velocity = vsin55 = Initial velocity of projectile in vertical direction = u

acceleration, a = 9.81 m/s²

displacement, s = 554 m

time, t = 8 s

Substituting,

554 = vsin55 x 8 + 0.5 x 9.81 x 8²

v = 36.64 m/s

Speed of the aircraft = 36.64 m/s