Atelevision camera at ground level is filming at the lift-off of a space shuttle that is rising vertically according to the position equation s = 50t 2, where s is measured in feet and t is measured in seconds. the camera is 2000 ft from the launch pad. what is the rate of change in the angle of elevation of the camera 10 seconds after lift-off?

2/29 rad/sec

Explanation: from the attachment ,

S=50t^2

S=50×10^2

S=5000

Tan¤= opp/Adj

=5000/2000

Tan-1 2.5=68.19°

Let y be the angle of elevation

Let S be the camera height

ds/dt= 50×2t= 100t

Substituting t=10

100(10)=1000

The rate of change of angle of elevation of the camera will be calculated using :

dy/dt=1/2000÷1+(5000/2000)2 ×1000

dy/dt=1/2÷(1+25/4)

dy/dt=2/29rad/sec

a person with a mass of 106kg has a weight of 1060n

probably not if net force is zero