Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water

See explanation below

Explanation:

This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.

First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:

V = V₀ + gt   (1)

X = V₀t + gt²/2   (2)

In this case, g = 9.8 m/s²

Now, let's see the displacement and velocity for each given time:

a) For  t = 0.5 s

V = 14 + (9.8)*0.5

V = 18.9 m/s

X = (14*0.5) + (9.8)(0.5)²/2

X = 7 + 1.225

X = 8.225 m

b) For t = 1.00 s

V = 14 + (9.8)*1

V = 23.8 m/s

X = (14*1) + (9.8)(1)²/2

X = 14 + 4.9

X = 18.9 m

c) For t = 1.5 s

V = 14 + (9.8)*1.5

V = 28.7 m/s

X = (14*1.5) + (9.8)(1.5)²/2

X = 21 + 11.025

X = 32.025 m

d) For t = 2 s

V = 14 + (9.8)*2

V = 33.6 m/s

X = (14*2) + (9.8)(2)²/2

X = 28 + 19.6

X = 47.6 m

e) For t = 2.50 s

V = 14 + (9.8)*2.5

V = 38.5 m/s

X = (14*2.5) + (9.8)(2.5)²/2

X = 35 + 30.625

X = 65.625 m

Hope this helps

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