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:
[tex] y = y_{0} + v_{o}t - \frac{1}{2}g*t^{2} [/tex] (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:
[tex] y = y_{0} + v_{o}t - \frac{1}{2}g*t^{2} [/tex] (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):
[tex] y_{watermelon} = y_{Superman} [/tex]
[tex] y_{0} + v_{o}t - \frac{1}{2}g*t^{2} = y_{0} + v_{o}t - \frac{1}{2}g*t^{2} [/tex]
[tex] 320 m + 0 - \frac{1}{2}9.8 m/s^{2}*t^{2} = 320 m - 36.0 m/s*t + 0 [/tex]
[tex] \frac{1}{2}9.8 m/s^{2}*t^{2} = 36.0 m/s*t [/tex] (3)
Solving equation (3) for t, we have:
[tex] t = 2*\frac{36.0 m/s}{9.8 m/s^{2}} = 7.35 s [/tex]
Hence, the watermelon reaches Superman in 7.35 s.
Now, we can finally find the speed of the watermelon when it passes Superman:
[tex] v_{f} = v_{0} - g*t [/tex]
Where:
vf: is the final speed of watermelon (when it passes Superman)
[tex] v_{f} = 0 - 9.8 m/s^{2}*7.35s = -72.03 m/s [/tex]
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!
