A plane, diving with constant speed at an angle of 55.0° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 55.0° with the vertical, releases a projectile at an altitude of 830 m. The projectile hits the ground 5.00 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?
_______ m/s

(b) How far did the projectile travel horizontally during its flight?
__________ m

(c) What were the horizontal and vertical components of its velocity just before striking the ground?
__________ m/s (horizontal)
__________m/s (vertical)

Explanation / Answer

Given data
Angle made by the plane with verticalis, = 55o Angle made by the plane with horizontal is, = 55o - 90o                                                                         = - 35o                                                                         = - 35o Projcectile released at an altitude , h = 830 m Time interval, t = 5.0 s Solution: (a) The speed of the plane is,                  - h = v0 sin t - 1/2 g t2              - 830 m = v0 sin ( -35o) (5.0 s) - 1/2 (9.8 m/s2 )(5.0 s) 2                       v0 = 247 m/s _____________________________________________________ _____________________________________________________                       v0 = 247 m/s _____________________________________________________ _____________________________________________________ (b) The horizontal distance travelled by the projectile is,                      x = v0 cos t                         = (247 m/s) cos (- 35o) (5.0 s)                         = 1011.65 m ______________________________________________________ ______________________________________________________ (c) Horizontal component of velocity is,                    vx = v0 cos(- 35o)                        = (247 m/s) cos(- 35o)                        = 202.33 m/s Vertical component of velocity is,                    vy =  v0 sin - gt                        =  (247 m/s) sin (- 35o) -(9.8 m/s2 )(5.0 s)                        = - 190.67 m/s downward                        =  (247 m/s) sin (- 35o) -(9.8 m/s2 )(5.0 s)                        = - 190.67 m/s downward

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