Let the angle of elevation is x and the height of the rocket from the ground is y
tanx = y/15
by differentiating both sides with respect to T
sec²x·dx/dt = (dy/dt)/15
at y = 30 , the hypotenuse of the triangle = 15√5
sec²x=(15√5/15)²=5
5 dx/dt = 11/15
dx/dt = 11/75 rad/sec
Answer:
yes its linear
Step-by-step explanation:
C should be the correct answer