Depends on the height of the boy throwing ;)

A solution please. Thank you!

If we ignore air resistance, object density and other factors, also if we assume that initial object height is 75 m + 2.2 m (boy's release height) = 77.2 m, than you need three kinematic equations to solve your tasks.

The answer (1) is 91.94 m (or 14.74 m from boy's release height) using kinematic equation (Vf^2 = Vi^2 + 2*a*d) + 77.2 m, where Vf is final velocity = 0; Vi - initial velocity = 17 m/s; a - acceleration (9.8 m/s) and d - distance. Time to reach highest height (code: HHt) is 1.73 s from second equation Vf = Vi + a * HHt.

The answer (2) is 6.06 s to reach bottom of the cliff ,using third kinematic equation (d = Vi*t + 1/2 * a * t^2) + HHt where d = 91.94 m; Vi = 0; a = 9.8 m/s.

Best regards.

@Darko Stojanović can I ask where or how did you get the boy's release height which is 2.2m?

To Rhea Mea Impas: According to this article:

Article Optimal release conditions for the free throw in men’s basketball

The free throw shooter was assumed to release the ball 2.134 m (7 ft) above the ground...

Two hints, Dear Rhea:

Use potential + kinetic energy conservation.

Set the reference system on the bottom of the cliff. It is easier.

Noted Sir Anton Vrdoljak and P Contreras. Thank You for your feedbacks very much appreciated.💗💗😊😊

You're welcome Rhea Mea Impas, I am glad that I was able to help...

My pleasure, Dear Rhea Mea Impas :

At the top of the cliff:

E = mgh1 + mv12/2

At the highest height:

E= mgh2 (v2 is zero)

Therefore:

mgh1 + mv12/2 = mgh2

solving for h2, h2 = h1 + v12/2g where h1 = 75 m, , v1 = 17 m/s and g = 9.8 m/s2, neglect boys height, it is much less than 75 m.

Then use the same equation for mechanical energy E, to find the final velocity

vf at position: - h1 or also called the bottom of the cliff:

At the botton of the cliff:

mgh2 = mvf2/2 since for the reference point h= 0, soving for velocity at the bottom gives finally, vf = sqrt(2gh2)

Finally, substitute the numbers and the units. and give some meaning of the values h2 and vf found. For example, the sign of the velocity vector at this point is negative.