A.) An inquisitive physics student and mountian climber climbs a 52.1 m cliff th

Question

A.) An inquisitive physics student and mountian climber climbs a 52.1 m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.35 s apart and observes that they cause a single splash. The first stone has an initial velocity of 2.2 m/s. How long after release of the first stone do the two stones hit the water?

B.) What is the initial velocty of the second stone if the two stones hit simultaniously?

C.) What is the speed of each stone when it hits the water, Enter the slower speed first.

Explanation / Answer

time taken by first stone
-using the second equation of motion
s= ut +1/2gt2

52.1= 2.2t+4.9t2

solving this we get

t=3.044s
so time taken by second stone= 3.044-1.35= 1.694s
putting it in the second eqn of motion
52.1= u*1.694+4.9*(1.694)2

u=22.455m/s

velocity of 2nd stone= 22.455m/s

(b) v2-u2= 2gs
so for the first stone u=2.2m/s
putting the value
v=32.031m/s

for 2nd stone u=22.455m/s
for 2nd stone v=39.056m/s
so velocity of 1st stone on hitting water=32.031m/s
so velocity of 2nd stone on hitting water=39.056m/s

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