A rock is tossed straight up with a speed of 22 m/s. When it returns, it falls i

Question

A rock is tossed straight up with a speed of 22 m/s. When it returns, it falls into a hole 14 m deep. In other words, assume that the rock lands 14 m lower than the height from which it was thrown. Take "up" to be the positive direction for this problem. Although the numbers in this problem have less than 3 significant figures, please enter your answers with at least 3 significant figures, and remember that velocity is a vector. (a) What is the rock's velocity as it hits the bottom of the hole? m/s (b) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Explanation / Answer

a)
d= v0*t +1/2*a*t^2 , where

-14 = -22*t + (0.5)*(-9.8)*t^2

=> t = 0.56 sec
Now, vfinal = vstart + a*t

So, v_f =-22m/s -9.8m/s*0.565sec
= -27.5 m/s (note that it's negative since it's going downward)

b)

t = v0/a = 22m/s/(9.8m/s^2) = 2.24s (to reach the top).

The time to come back down to ground level is the same (2.24s)

The total time before hitting the bottom of the hole is just the sum of these:

t = 2.24s + 2.24s + 0.565s = 5.04 seconds

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