Question

A plane, diving with constant speed at an angle of 51.0

Answer 1

The initial speed of the plane, V0, at 51 degrees angle with the vertical can be decomposed into 2 components:
The horizontal speed: Vx=V0*sin(51)
The vertical speed: Vy=V0*cos(51)

(a)
For the vertical motion, the projectile drops 620m in 8 sec,
h=(1/2)*g*t^2 + Vy*t
or
h=(1/2)*g*t^2 + V0*cos(51)*t

with h=620, g=9.8, t=8, the only unknown is V0.

Substituting these into the equation and solving for V0,
620 = (1/2)*(9.8)*(8^2) + V0*(cos(51))*8

V0 = 60.89 = 60.9 m/s
(b) The horizontal motion is a constant motion so the distance it travels is
d = Vx * t
or
d = V0 * sin(51) * t
d = 60.9 * sin(51) * 8
d = 378.49 m

(c)
For the vertical motion find the vertical speed, Vyf, at which the projectile hits the ground using
Vyf = g*t + Vy
Vyf = g*t + V0*cos(51)
Vyf = (9.8)*(8) + 60.9(cos(51))
Vyf = 116.69 m/s (vertical speed)

For the horizontal motion, it is contant speed with
Vx = V0 * sin(51)
Vx = 60.89 * sin(51)
Vx = 47.31 m/s (horizontal speed)