Question

a rock is thrown upward from a bridge that is 72 ft above a road. The rock reaches its maximum height above the road 0.97 seconds thrown and contacts the road 3.84 seconds after it was thrown. write a function f that determines the rocks height above the road (in feet) in terms of the numbers of seconds t since the rock was thrown

Answer 1

The rock reaches the maximum height in 0.97 seconds

At maximum height final velocity is zero.

V = u - gt

0 = u - 9.81 x 0.97

u = 9.5157 m/s

u = 31.2 ft/s

g = 9.81 m/s² = 32.2 m/s²

H_road = H_bridge + ut - ½gt²

H_road = 72 + 31.22t - 16.1t²

Verification

The rock will reach reach H_max (from bridge) at 0.97 s. The rock will come back to bridge height 72 ft at a time period 2 x 0.97 = 1.94 s

H_road = 72 + 31.22t - 16.1t²

72 = 72 + 31.22(1.94) - 16.1(1.94)²

72 = 72 + 60.6 - 60.6

72 = 72