An inquisitive physics student and mountain climber climbs a 52.0 m cliff that o

Question

An inquisitive physics student and mountain climber climbs a 52.0 m cliff that overhangs a calm pool of water. He throws (not drops) two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.05 m/s downward. Hint: Because the motion is only in the downward direction, choose the positive y-axis to point downward.

(a) How long after release of the first stone do the two stones hit the water?

(b) What initial velocity must the second stone have if they are to hit simultaneously?(downward)

(c) What is the speed of each stone at the instant the two hit the water?
first stone (downward)

second stone (downward)

Explanation / Answer

Let choose the positive y-axis to point downward.

Y = ut + 1/2gt2

for stone one

52 = 2.05t + 1/2 g t2

by solving this we get t = 3sec

b)

time of flight of second stone is 3-1 = 2sec

52 = u*2 + 1/2*9.8 *4

u = 32.4/2 = 16.2 m/s

C)

V1 = 2.05 +9.8*3 = 31.45m/s

V2 = 16.2 + 9.8*2 = 35.8 m/s

52 = u(t-1) + 1/2 g*(t

2

eq.2- eq.1

0= u(t-1)-2.05t + 1/2 g *( -2t-1)

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