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
