Question

For this problem, a person standing on the edge of a cliff throws a rock directly upward. it is observed that 2 secs later the rock is at its maximum height (in feet) and that 5 secs after that (meaning 7 seconds after being thrown) the rock hits the ground at the base of the cliff.

(a) What is the initial velocity of the rock? ft/sec.

(b) How high is the cliff? t ft.

(c) What is the velocity of the rock at time t? ft/sec.



(d)What is the velocity of the rock when it hits the ground? ft/sec.

Answer 1

a. V = Vo + g*t = 0
Vo - 32*2 = 0
Vo = 64 Ft/s.

b. Tf1 = Tr = 2 s. = Fall time from hmax
to top of cliff.

Tf2 = 7-Tr-Tf1 = 7 - 2 - 2 = 3 s. = Fall
time from top of cliff to gnd.

h = Vo*Tf2 + 0.5g*Tf2^2
h = 64*3 + 16*3^2 = 336 Ft. = Ht. of cliff.

c. What is time t?

d. h max = ho + -(Vo^2)/2g
h max = 336 + -(64^2)/-64 = 400 Ft. Above gnd.

V^2 = Vo^2 + 2g*h = 0 + 64*400 = 25,600
V = 160 Ft/s.