Question

As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 24 m/s. How fast is the watermelon going when it passes Superman?

Answer 1

Answer:

47.9709 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

For watermelon

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow s=\frac{1}{2}\times 9.81\times t^2\ m$

For superman

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times t+\frac{1}{2}\times 0\times t^2\\\Rightarrow s=24t\ m$

The distance travelled by the watermelon and superman will be same

$\frac{1}{2}\times 9.81\times t^2\ m=24t\\\Rightarrow t=\frac{24}{0.5\times 9.81}\\\Rightarrow t=4.89\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 4.89\\\Rightarrow v=47.9709\ m/s$

The watermelon is going at a speed of 47.9709 m/s when it passes Superman

Answer 2

The velocity of the watermelon when it passes Superman is 48 m/s.

The given parameters;

• height above the ground the watermelon was released, h = 320 m

• velocity of superman, v = 24 m/s

Let the time the watermelon passes superman = t

• The distance moved by Superman = 24t

• The distance traveled by the watermelon = $\frac{1}{2} gt^2$

Solve the two equations together to obtain the value of t;

$\frac{1}{2} gt^2 = 24t\\\\gt = 48\\\\t = \frac{48}{9.8} \\\\t = 4.9 \ s$

The velocity of the watermelon at this time is calculated as;

$v_f = v_0 + gt\\\\v_f = 0 \ + \ 48 \\\\v_f = 48 \ m/s$

Thus, the velocity of the watermelon when it passes Superman is 48 m/s.

Learn more here: https://brainly.com/question/13372763