1 . NOTE: For this problem, WebAssign defines the downward direction as negative

Question

1 . NOTE: For this problem, WebAssign defines the downward direction as negative. Since velocity is a vector, you will have to indicate the direction by the sign of your answer, as either positive or negative.
A ball is thrown straight down and takes 1.50 s to travel (fall) a distance of 15.5 m.

Calculate the initial velocity of the ball. Your answer should have 3 significant figures.

AND

A rock is tossed straight up with a speed of 22 m/s. When it returns, it falls into a hole 12 m deep. In other words, assume that the rock lands 12 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?
m/s

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

Explanation / Answer

here,

time taken by the ball , t = 1.5 s

distance travelled by ball ,s = 15.5 m

(a)

let the initial velocity be u

using 2nd equation of motion,

s = u * t + 0.5 * a * t^2

15.5 = u * 1.5 + 0.5 * 9.8 * 1.5^

u = 2.983 m/s

the initial velocity of the ball is - 2.983 m/s

(b)

for t = 0.9 s

let the velocity of ball be v

using first equation of motion

v = u + a*t

v = 2.983 + 9.8 * 0.9

v = 11.803 m/s

the velocity of ball after t = 0.9 is - 11.803 m/s

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