Question

a person standing on the edge of a high cliff throws a rock straight up with an initial velocity v0 of 13 m/s. calculate the position and velocity of the rock at 1.00 s.

Answer 1

H = V0 t - 1/2 g t^2 since V0 and g are in different directions

H = 13 * 1 - 1/2 * 9.80 * 1 = 13 - 4.9 = 8.1 m

The rock is 8.1 m above its starting point after 1 second

V = V0 - g t = 13 - 9.8 * 1 = 3.2 m/s positive after 1 second