contestada

A rock takes 2.95 s to hit the ground when thrown straight up from a cliff with an initial velocity of 7.5 m/s.
a. Calculate the height of the cliff.
b. Determine the acceleration due to gravity.
c. Find the final velocity of the rock before hitting the ground.
d. Identify the time it takes to reach the maximum height.

Respuesta :

Explanation:

Position is given by the standard equation

d =  vo t - 1/2 at^2     where d is distance from the start point

d = 7.5 (2.95) - 1/2 (9.81)(2.95)^2

d = - 20.6 m      so the height is   20.6 meters

b)  acceleration due to gravity is a constant =    9.81 m/s^2  ( downward)

c) v f = vo  + a t

    vf = 7.5  + 9.81 ( 2.95) = -21.44 m/s     ( or 21.44 m/s downward)

d) at max height v = 0

    vf = vo + at

     0 = 7.5 - 9.81 t       t = .76 sec

Otras preguntas