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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?

Respuesta :

Answer:

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

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