A rock is tossed straight up with a speed of 18 m/s. When it returns, it falls i

Question

A rock is tossed straight up with a speed of 18 m/s. When it returns, it falls into a hole 13 m deep. In other words, assume that the rock lands 13 m lower than the height from which it was thrown. Take "up" to be the positive direction for this problem. Although the numbers in this problem have less than 3 significant figures, please enter your answers with at least 3 significant figures, and remember that velocity is a vector.

a) What is the rock's velocity as it hits the bottom of the hole?


(b) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Explanation / Answer

Here ,

u = 18 m/s

vertical displacement , y = - 13 m

a) Using third equation of motion

v^2 - u^2 = 2 * a * d

v^2 - 18^2 = 2 * 9.8 * 13

v = -24.1 m/s

the rock's velocity is - 24.1 m/s

b)

let the time taken is t

v = u + a * t

-24.1 = 18 - 9.81 * t

t = 4.29 s

the time taken is 4.29 s

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