A plane, diving with a constant speed at an angle of 48.0 degree from the vertic

Question

A plane, diving with a constant speed at an angle of 48.0 degree from the vertical, releases 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. What is the speed of the aircraft? How far did the projectile travel horizontally during its flight? What was the projectile's velocity in unit vector notation just before striking the ground?

Explanation / Answer

the angle of the initial velocity vector with the positve horizontal xaxis is =-48o, and the initial height is y0=690m.
We know that at t=8s, the projectile is on the gorund, which we will take as y=0.
The equation for the ycomponent of the position vector is

y = y0 + v0sint - 1/2gt^2

so at t=6s, when y=0, we know every thing but the initial speed:

0 = y0 + v0sint - 1/2gt^2

=> v0 = 1/2gt^2 - y0 / sint

= (0.5*9.8*(8)^2 - 690 ) / ((sin(-48) *8 )

= 63.3120 m

b)
Since we know the projectile travels with constant velocity in the horizontal direction, we know that distance = velocity * time:

d = v0x * t
= v0cost
= 63.3120*cos(-48)*8
= 338.912 m

c) We need x and y components of the velocity at t=6s:

vx = v0cos = 63.3120*cos(-48) = 42.36 m/s
vy = v0sin - gt = 63.3120*sin(-48) - 9.8*8 = -125.45 m/s
v = sqrt(vx^2 + vy^2)
= 132.41 m/s

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