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 36.0 m/s.
How fast is the watermelon going when it passes Superman?

Answer 1

Answer:

-72.03 m/s (since it's moving down).

Explanation:

To find the speed of the watermelon going when it passes Superman, we need to see both cases, the watermelon and Superman:

For Watermelon:  

$y = y_{0} + v_{o}t - \frac{1}{2}g*t^{2}$    (1)

Where:

y₀ = is the initial height = 320 m

v₀ = is the initial speed = 0 (since it's dropped)  

g = is the gravity = 9.8 m/s²      

t: is the time

For Superman:  

$y = y_{0} + v_{o}t - \frac{1}{2}g*t^{2}$   (2)

Where:

y₀ = is the initial height = 320 m      

v₀ = is the initial speed = -36.0 m/s                

g = is the gravity = 0 (since he's flying at speed constant)  

t: is the time

Since the height, y, is the same for watermelon and Superman when they meet (when the watermelon reaches Superman), we can equal equations (1) and (2):        

$y_{watermelon} = y_{Superman}$    

$y_{0} + v_{o}t - \frac{1}{2}g*t^{2} = y_{0} + v_{o}t - \frac{1}{2}g*t^{2}$

$320 m + 0 - \frac{1}{2}9.8 m/s^{2}*t^{2} = 320 m - 36.0 m/s*t + 0$    

$\frac{1}{2}9.8 m/s^{2}*t^{2} = 36.0 m/s*t$  (3)

Solving equation (3) for t, we have:

$t = 2*\frac{36.0 m/s}{9.8 m/s^{2}} = 7.35 s$

Hence, the watermelon reaches Superman in 7.35 s.

Now, we can finally find the speed of the watermelon when it passes Superman:

$v_{f} = v_{0} - g*t$

Where:

vf: is the final speed of watermelon (when it passes Superman)

$v_{f} = 0 - 9.8 m/s^{2}*7.35s = -72.03 m/s$

Therefore, the speed of the watermelon when it passes Superman is -72.03 m/s. The minus sign is because the watermelon is moving down.

I hope it helps you!