A.) Calculate the maximum height of the stone above the cliff top. B.) Calculate the time which elapses before the stone hits the beach. C.) Calculate the speed with which the stone hits the beach.

Question

A boy stands on the edge of a cliff of the height of 80m. He throws a stone vertically upwards so that its distance, hm, above the cliff top is given by h=30t-5t^2.

Answer 1

set h = 0, where 0 indicates ground level or beach level.

h = -5t^2 + 30t + 80 = 0

divide by -5

t^2 -6t - 16 = 0

factor

(t-8)(t+2) = 0

set t-8=0

t = 8 seconds for the rock to reach the beach

max height is when -10t +30 = 0 or t=3 seconds

max height = -5(3)^2 +30(3) + 80 = -45+90+80= 125 meters high or 125-80 = 55 meters above the cliff.

in 6 seconds the rock is 80 meters high, back where it started, at the cliff level

in 2 more seconds it hits the beach

velocity = h'(t) = -10t +30

h'(8) = -10(8) +30 = -80+30 = -50 meters per second = speed when it hits the beach. the negative speed means it's going down.