From the top of a cliff, a person throws a stone straight downward. The initial

Question

From the top of a cliff, a person throws a stone straight downward. The initial speed of the stone just after leaving the person's hand is 9.8 m/s. (a) What is the acceleration (magnitude and direction) of the stone while it moves downward, after leaving the person's hand? magnitude 9.8 direction downward m/s2 Is the stone's speed increasing or decreasing? increasing decreasing (b) After 0.51 s, how far beneath the top of the cliff is the stone? (Give just the distance fallen, that is, a magnitude.) Be careful of signs. If the downward direction is negative, and the initial velocity and acceleration are both downward, what are their algebraic signs? Think about what information you are given, and which constant acceleration relation will allow you to find the distance. m

Explanation / Answer

B)

Using 2nd equation of motion

D= ut+05gt^2

= 9.8*0.51+0.5*9.8*0.51^2

= 6.272 m

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Comment in case any doubt.. good luck

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