A plane, diving with constant speed at an angle of 53.0 degree with the vertical

Question

A plane, diving with constant speed at an angle of 53.0 degree with the vertical, releases a projectile at an altitude of 500 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.) What is the speed of the aircraft? How far did the projectile travel horizontally during its flight? What were the horizontal and vertical components of its velocity just before striking the ground?

Explanation / Answer

a)

here by using the second equation of motion

y - y0 = uy * t + 0.5 * a * t^2

0 = y0 + uy * t + 0.5 *a * t^2

0 = 500 - u * cos53 * 6 - 4.9 * 6^2

u = (4.9 * 6 * 6 - 500 ) / ( 6 * -cos53)

u = 90 m/s

b)

use first equation of motino

x = vx * t

x = 90 * sin53 * 6

x = 431 m

c)

vx = 90 * sin53

vx = 71.8 m/s

d)

vy = uy + a * t

vy = -90 * cos53 - 9.8 * 6

vy = 113 m/s

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