A plane, diving with constant speed at an angle of 46.0° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 46.0° with the vertical, releases a projectile at an altitude of 560 m. The projectile hits the ground 7.00 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? (Indicate the direction with the sign of your answer.)

Explanation / Answer

Here, the initial speed of the plane, V0, at 46 degrees angle with the vertical can be decomposed into 2 components:
The horizontal speed: Vx=V0*sin(46)
The vertical speed: Vy=V0*cos(46)

(a) For the vertical motion, the projectile drops 560m in 7 sec,
h=(1/2)*g*t^2 + Vy*t
or
h=(1/2)*g*t^2 + V0*cos(46)*t

with h=560, g=9.8, t=7, the only unknown is V0.

Substituting these into the equation and solving for V0,
560 = (1/2)*(9.8)*(7^2) + V0*(cos(46))*7

=> 560 = 240.1 + 4.86* V0

=> V0 = 65.82 m/s

(Please note that V0 has an angle 46 degrees with the vertical)

(b) The horizontal motion is a constant motion so the distance it travels is
d = Vx * t
or
d = V0 * sin(46) * t
d = 65.82 * sin(46) * 7
d = 331.43 m

(c) For the vertical motion find the vertical speed, Vyf, at which the projectile hits the ground using
Vyf = g*t + Vy
Vyf = g*t + V0*cos(46)
Vyf = (9.8)*(7) + 65.82*(cos(46))
Vyf = 114.32 m/s (vertical speed)

For the horizontal motion, it is contant speed with
Vx = V0 * sin(46)
Vx = 65.82 * sin(46)
Vx = 47.35 m/s (horizontal speed)

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