A camera, located 4 km from the launch pad, is tracking the rocket that is fired straight up. When the height of the rocket is 19 km, the camera is rotating at the rate of 5/361 radians per second. What is the speed of the rocket at that instant? Give your answer in km/sec.
2006-11-04 06:44:02 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
h = 4tanθ
Vh = dh/dt = (dh/dθ)(dθ/dt)
dh/dθ = 4/cos^2θ
cos^2θ = 4^2/(4^2 + 19^2)
Vh = (16 + 361)/4)(5/361)
Vh = (5/4)(16/361 + 1)
Vh = 1.3054 km/sec
2006-11-04 07:34:11 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
let draw a circle with center the camera and the rocket on the circle. the radius of the circle is sqrt(19²+4²)=19.42km
We can approximate the distance on the circle by r*theta as theta is small with theta the angle so 19.42*5/361 = 0.269 km
So the speed is approximatively 0.27 km/sec
2006-11-04 15:34:48 · answer #2 · answered by fred 055 4 · 0⤊ 0⤋