Question

Assume air resistance is negligible unless otherwise stated. Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial velocity of 10.0 m/s from the Verrazano Narrows bridge in New York City. The roadway of this bridge is 70.0 m above the water. (Enter the magnitudes.)

Answer 1

Answer: 1) t=0.5 s; S=6.225 m and v=14.9 m/s

2) t=1 s; S=14.9 m and v=19.8 m/s

3) t=1.5 s; S=26.05 m and v=24.7 m/s

Explanation:

The displacement $S$ is given by

$S=ut+\frac{1}{2} at^{2}$

and  final velocity $v$ is given by

$v=u+at$

where $u$ is the initial velocity

$a$ is acceleration

$t$ is time taken

Case 1: when time is 0.5 s

The displacement is

$S=ut+\frac{1}{2} at^{2}\\S=10\times 0.5 +\frac{1}{2}\times 9.8\times 0.5^{2}\\\\S=6.225 m$

the velocity is

$v=u+at\\v=10+9.8\times 0.5\\v=14.9 m/s$

Case 2: when t=1 sec

The displacement is

$S=ut+\frac{1}{2} at^{2}\\S=10\times 1 +\frac{1}{2}\times 9.8\times 1^{2}\\\\S=14.9 m$

the velocity is

$v=u+at\\v=10+9.8\times 1\\v=19.8 m/s$

Case 3: t=1.5 s

The displacement is

$S=ut+\frac{1}{2} at^{2}\\S=10\times 1.5 +\frac{1}{2}\times 9.8\times 1.5^{2}\\\\S=26.05 m$

the velocity is

$v=u+at\\v=10+9.8\times 1.5\\v=24.7 m/s$