An astronaut throws a rock straight up from a cliff on a newly discovered planet. The height of the cliff is 12.16 m. The rock is thrown with an initial velocity of 8.2 m/s and hits the ground below the cliff 2.65 seconds later. y (m) x (m) What is the height of the rock, measured relative to the cliff top, at its highest point? (Note: Because the astronaut is not on Earth, the acceleration in the y-direction is not equal to −.)

The height of the rock, measured relative to the cliff top, at its highest point = 18.68 m

The height of the rock, measured relative to the ground, at its highest point = 30.84 m

Explanation:

The rock is thrown with an initial velocity, u = 8.2 m/s.

a = - 1.62 m/s² (negative sign because the rock is moving upwards against the planet's gravity)

At the maximum height reached, velocity, v = 0 m/s

y = maximum height reached, measured from the top of the cliff.

Using the equations of the motion

v² = u² + 2ay

0 = 8.2² + 2(-1.8)y

- 3.6y = - 67.24

y = 18.68 m

So the maximum height reached by the rock, measured from the ground = 18.68 + 12.16 = 30.84 m

ok

explanation: