A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls i

Question

A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls into a hole13 m deep. (a) What is the rock's velocity as it hits thebottom of the hole?
1 m/s (downward)
(b) How long is the rock in the air, from the instant it isreleased until it hits the bottom of the hole?
2 s

2)
A skier is gliding along at 3.0 m/s onhorizontal, frictionless snow. He suddenly starts down a6° incline. His speed at thebottom is 15 m/s. (a) What is the length of the incline?
1 m
(b) How long does it take him to reach the bottom?
2 s


(a) What is the rock's velocity as it hits thebottom of the hole?
1 m/s (downward)
(b) How long is the rock in the air, from the instant it isreleased until it hits the bottom of the hole?
2 s (a) What is the length of the incline?
1 m
(b) How long does it take him to reach the bottom?
2 s

Explanation / Answer

a) the max height is: h = v2/2g = (20m/s)2/ (2*9.8m/s2) = 20.41m at max height v = 0 now the rock will fall the distance H = h + 13m = 33.41m the time it takes is t, and H = (1/2)gt2, =>t = (2*H/g) = (2*33.41m/9.8m/s2)= 2.61 s the speed at the bottom: vf = gt =9.8m/s2*2.61s = 23.49 m/s (downward) b) the rock to reach max height is t1 = v/g = 20m/s /9.8m/s = 2.04s the total time = t+t1 = 2.61s + 2.04s = 4.65s

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