Answer:
v = 30.39 m/s
Explanation:
given,
mass of glider,M = 680 Kg
mass of the skydiver, m = 68 Kg
horizontal velocity,V = 30 m/s
when skydiver releases, velocity,v = 30 m/s
velocity if the glider,v' = ?
use the conservation of momentum
M V = m' v' + m v
m' = 680-68 = 612 Kg
680 x 30 = 612 x v + 60 x 30
612 v = 18600
v = 30.39 m/s
since the skydiver's speed will be the same as before release
The glider should continue to travel at 30.39 m/s since there are no external forces acting on it
Kepler's first law states that planetary orbits are elliptical in shape. (C)
Answer:
a ) 540 ft /s
b ) .144° /s
Explanation:
Let at any moment h be the height of the rocket . The distance of rocket from camera will be R
R ² = 4000² + h²
Differentiating both sides with respect to t
R dR/dt = h dh/dt
dR/dt = h/R dh/dt
a ) Given speed of rocket
dh/dt = 900
h = 3000
R² = 4000² + 3000²
R = 5000
dR/dt = h/R dh/dt
= (3000 / 5000 ) X 900
dR/dt = 540 ft /s
b ) Let θ be angle of elevation at the moment .
4000 / R = cosθ
Differentiating with respect to t
- 4000 x 1 / R² dR/dt = - sinθ dθ / dt
4000 x ( 1/ 5000² ) x 540 = 3 /5 x dθ / dt
.0864 = 3/5 dθ / dt
dθ / dt = .144° /s
