Question

A plane, diving with constant speed at an angle of 44.3° with the vertical, releases a projectile at an altitude of 667 m.The projectile hits the ground 6.80 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

Here ,

let the speed of plane is v

a) time , t = 6.80 s

Using second equation of motion

y = ut + 0.5at^2

667 = u * sin(44.3) * 6.80 + 0.5 9.8 * 6.38^2

solving for u

u = 98.44 m/s

the speed of plane is 98.44 m/s

b)

distance = v * cos(theta) * t

distance = 98.44 m/s* cos(44.3) * 6.80

distance = 479 m

the plane travel 479 m horizontally

c)

as the hoizontal component is same

Vh = 98.44 m/s* cos(44.3)

Vh = 70.45 m/s

d)

Vy = V * sin(theta) + 6.8 * 9.8

Vy = 98.44 * sin(44.3) + 6.8 *9.8

Vy = 135.4 m/d

tr vertical speed is 135.4 m/s