Answer:
height of the cliff = 74.529 meters
Explanation:
in this problem we will write projectile motion equations in x and y directions.
writing the equation of motion
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
where s is the distance traveled in x or y direction, u the initial velocity in x or y direction, t is the time travelled, a is the acceleration along x or y direction.
therefore along y direction [tex]u_{y}[/tex] = 0 and [tex]a_{y}[/tex] = -9.8 m/[tex]s^{2}[/tex] .
therefore along y direction the equation becomes [tex]H = -\frac{1}{2}gt^{2}[/tex] applying the values [tex]H = -\frac{1}{2}9.8\times 3.9^{2}= 74.529meters[/tex]
