Answer:
16.45 m/s
Explanation:
Let y be the vertical distance and x be the horizontal distance
We are given that
The altitude of hang glider increasing at the rate=[tex]v_y=\frac{dy}{dt}==6.75m/s[/tex]
The shadow of the glider moves along the ground at speed=[tex]v_x=\frac{dx}{dt}=15m/s[/tex]
We have to find the magnitude of glider's velocity.
We know that
Magnitude of velocity=[tex]v=\sqrt{v^2_x+v^2_y}[/tex]
Substitute the values
[tex]v=\sqrt{(15)^2+(6.75)^2}[/tex]
[tex]v=\sqrt{225+45.5625}[/tex]
[tex]v=16.45m/s[/tex]
Hence, the magnitude of glider's velocity=16.45 m/s
