Question

A plane, diving with constant speed at an angle of 51.5° with the vertical, releases a projectile at an altitude of 654 m. The projectile hits the ground 6.40 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.)

Answer 1

h= v(oy) +gt²/2
v(oy)=( h/t )– (gt/2) = 7.83 m/s.

v(oy) =v(o) •cosα .
(a)
v(o) = v(oy)/cosα = 70.83/cos51.5 = 113.8 m/s.
(b) v(ox)=v(x) = v(o)sinα=
=113.8•sin51.5 =89.06 m/s
s=v(x) •t =89.06•6.4=570 m.
(c)
v(y) =v(oy) +gt =7.83 +9.8•6.4=70.55 m/s.
v(x) =89.06 m/s