4. A plane, diving with a constant speed at an angle of 48.0° from the vertical,

Question

4. A plane, diving with a constant speed at an angle of 48.0° from the vertical, releases a projectile at an altitude of 690 m. The projectile hits the ground 8.00 s after release. Assume a coordinate system in which the airplane is moving in the positive horizontal direction and the negative vertical direction. Neglect air resistance. a. What is the speed of the aircraft? b. How far did the projectile travel horizontally during its flight? c. What was the projectiles velocity in unit vector notation just before striking the ground?

Explanation / Answer

The initial speed of the plane, V0, at 48 degrees angle with the vertical can be decomposed into 2 components:
The horizontal speed: Vx=V0*sin(48)
The vertical speed: Vy=V0*cos(48)

(a)
For the vertical motion, the projectile drops 760m in 4 sec,
h=(1/2)*g*t^2 + Vy*t
or
h=(1/2)*g*t^2 + V0*cos(48)*t

with h=690, g=9.8, t=8, the only unknown is V0.

Substituting these into the equation and solving for V0,
690 = (1/2)*(9.8)*(8^2) + V0*(cos(48))*8

V0 would be 70.31 m/s
(this V0 has an angle 48 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(48) * t
d = 70.31 * sin(48) * 8
d = 418 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(48)
Vyf = (9.8)*(8)*(cos(48))
Vyf = 52.45 m/s (vertical speed)

For the horizontal motion, it is contant speed with
Vx = V0 * sin(48)
Vx = 70.31 * sin(48)
Vx = 52.25 m/s (horizontal speed)

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