Falling Objects

Question

(a)

Calculate the height (in m) of a cliff if it takes 2.15 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.05 m/s. (Enter a number.)

 m

(b)

How long (in s) would it take to reach the ground if it is thrown straight down with the same speed? (Enter a number.)

 s

Answer 1

For part (a), use the kinematics formula d = v_it + 1/2 at², with v_i = 8.05 m/s, t = 2.15 s, and a = g = -9.8 m/s². With these substitutions, you will find that the displacement of the rock during the 2.15 s equals - 5.34 m, indicating that the height of the cliff is 5.34 m.

For part (b), use the same formula. This time, the displacement is known (same as in (a), since it is the same cliff), acceleration is again g (-9.8 m/s²), and v_i equals - 8.05 m/s (same speed as in (a), but down). Substituting these values into the equation d = v_it + 1/2 at², will give you a quadratic expression. The positive value of t (just over half a second) is the answer you are looking for.

-5.34 = -8.05 t - 4.9 t² 4.9 t² + 8.05 t - 5.34 = 0 t =( -8.05 +/- (8.05² - 4*-5.34*4.9)^1/2) / (2 * 4.9)