Your response is within 10% of the correct value. This may be due to roundoff er

Question

Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s An inquisitive physics student and mountain climber climbs a 54.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.04 m/s. How long after release of the first stone do the two stones hit the water? What initial velocity must the second stone have if the two stones are to hit the water simultaneously? magnitude m/s direction What is the speed of each stone at the instant the instant the two stones hit the water? first stone m/s second stone m/s

Explanation / Answer

using laws of motion , this problem can be solved

(a) h = u*t + g*t2/2

54= 2.04 *t + 9.81 *t2/2

On solving we get time (t)= 3.116 seconds.

(b) given that 2 nd stone is thrown a sec later, which means time taken for the 2 stone to touch the water surface = 2.116 seconds.

Using the equation from (a) : 54= u2*2.116 + 9.81 * 2.1162/2, we get u2= 15.14 m/s

(c) Velocity with which stones strike the surface is given by : v2-u2= 2*g*h

for stone 1: v1= square root (2.042 + 2*9.81*54) =32.61 m/s

for stone 2: v2= square root (15.142 + 2*9.81*54) =35.898 m/s.

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