An initial velocity of the projectile v₀ = the velocity of the plane.
v₀ (x) = v₀cosα,
v₀ (y) = v₀sinα,
h= v₀ (y) •t+ gt²/2=
=v₀sinα•t + gt²/2.
v₀={h- gt²/2}/ sinα•t=
={790 – (9.8•36/2)}/sin52•6 =129 m/s
s= v(x)t= v₀cosα•t=790•cos52•6=479.4 m.
v(x) = v₀ (x) = v₀cosα= 129 •cos52 =79.4 m/s
v(y) = v₀ (y)+gt = v₀sinα +gt =
=160.5 m/s
A plane, diving with constant speed at an angle of 52.0° with the vertical, releases a projectile at an altitude of 790 m. The projectile hits the ground 6.00 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?
(b) How far did the projectile travel horizontally during its flight?
(c) What were the horizontal and vertical components of its velocity just before striking the ground?
1 answer
1 answer