An airplane diving at an angle of 50.0 degrees with the vertical releases a miss

Question

An airplane diving at an angle of 50.0 degrees with the vertical releases a missile at an altitude of 750 m. The missile hits the ground 5.00 s after it is released. (a) What is the speed of the aircraft? (b) What is the horizontal distance traveled by the missile before it hits the ground?(c) Sketch the following two graphs: i) |vy| vs t, ii) y vs t. (Note: The origin of both graphs should be (0,0). The graphs need not be to scale, but you should give the coordinates of the points when t = 0 and t = 5 s.)

Explanation / Answer

The initial speed of the plane, V0, at 50 degrees angle with the vertical can be decomposed into 2 components:
The horizontal speed: Vx=V0*sin(50)
The vertical speed: Vy=V0*cos(50)

(a)
For the vertical motion, the projectile drops 750m in 5 sec,
h=(1/2)*g*t^2 + Vy*t
or
h=(1/2)*g*t^2 + V0*cos(50)*t

with h=750, g=9.8, t=5, the only unknown is V0.

Substituting these into the equation and solving for V0,
750 = (1/2)*(9.8)*(5^2) + V0*(cos(50))*5

V0 would be 195 m/s (check my math)
(this V0 has an angle 50 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(50) * t
d = 195 * sin(50) * 5
d = 746 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(50)
Vyf = (9.8)*(5)*(cos(50))
Vyf = 31.5 m/s (vertical speed)

For the horizontal motion, it is contant speed with
Vx = V0 * sin(55)
Vx = 195 * sin(50)
Vx = 149.4 m/s (horizontal speed)

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