Question

Please helpppppppppp

Answer 1

The height of the cliff is 96.2 m

Explanation:

The motion of the rock along the horizontal direction is a uniform motion (constant velocity), since there are no forces acting along that direction. Therefore, the horizontal velocity is given by

$v_x = \frac{d}{t}$

where, using the data of the first rock:

d = 66.5 m is the horizontal distance travelled by the rock

$v_x = 15 m/s$ is the horizontal velocity

Solving for t, we find the time it takes for the rock to reach the ground:

$t=\frac{d}{v_x}=\frac{66.5}{15}=4.43 s$

The vertical motion instead is a free fall motion acted upon gravity, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$  

where  

s is the vertical displacement of the rock, which corresponds to the height of the cliff

u = 0 is the initial vertical velocity of the rock

t = 4.43 s is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity  (we chose downward as positive direction)  

Solving for s, we find the height of the cliff:

$s=0+\frac{1}{2}(9.8)(4.43)^2=96.2 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly