a stone is thrown vertically upward with a speed of 10.5 m/s from the edge of a cliff 76.0 m high.

Question

how much later does it reach the bottom of cliff?
what is its speed just before hitting?
what total distance did it travel

Answer 1

V = Vo + g*Tr = 0
10.5 - 9.8*Tr = 0
9.8t = 10.5
Tr = 1.07 s. = Rise time.

hmax = ho + (V^2-Vo^2)/2g =
76 + (0-10.5^2)/-19.6 = 81.63 m. Above gnd.

hmax = Vo*t + 0.5g*t^2 = 81.63 m.
0 + 9.8*t^2 = 81.63
t^2 = 8.33
Tf = 2.89 s. = Fall time.

Tr+Tf = 1.07 + 2.89 = 3.96 s. To reach
bottom of cliff.

V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*81.63 = 1599.95
V = 40 m/s Just before hitting gnd.

d = (hmax-ho)+ hmax
d = (81.63-76) + 81.63 = 87.26 m. =
Total distance traveled.