A plane, diving with constant speed at an angle of 52.0° with the vertical, rele

Question

A plane, diving with constant speed at an angle of 52.0° with the vertical, releases a projectileat an altitude of 790 m. The projectilehits the ground 4.00 s after release.(Assume a coordinate system in which the airplane is moving in thepositive horizontal direction, and in the negative verticaldirection. Neglect air resistance.)
(a) How far did the projectile travel horizontally during itsflight?

(b) What were the horizontal and vertical components of itsvelocity just before striking the ground?

please explain all steps. and solve. Thanks

(a) How far did the projectile travel horizontally during itsflight?

(b) What were the horizontal and vertical components of itsvelocity just before striking the ground?

please explain all steps. and solve. Thanks

Explanation / Answer

h = vyi t + 1/2 g t2   wherevyi is the initial vertical downward speed of projectile vyi = 178   after substituting thegiven values vyi = V sin 38 where V = 289 m/s =speed of plane vxi = vx = constant = V cos 38 = 228m/s vyf = vyi + g t = 178 + 9.8 * 4 =217 m/s This gives the horizontal and vertical components ofvelocity Sx = vx t = 228 * 4 = 912 m

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