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 constant speed of 35 m/s. How fast is the watermelon going when it passes Superman

Answer 1

70 m/s

Step-by-step explanation:

The watermelon will go pass superman at the instance when they are at the same distance from the top of the building.

Superman flies by with a constant velocity = 35 m/s

This distance where he will be overtaken by the watermelon will be = vt

where v is his velocity

t is the time the watermelon and superman are side by side

For the watermelon,

the initial velocity u = 0 m/s (since it starts from rest)

The distance where it is side by side with superman will be gotten from the equation

s = ut +
`(1)/(2)gt^2`

where is the distance down at which they meet

g is the acceleration due to gravity = 9.81 m/s^2

t is the time they meet

ut = 0, since u = 0 m/s

Equating this distance that they meet, we'll have

vt =
`(1)/(2)gt^2`

substituting values, we have

35t =
`(1)/(2)*9.81*t^2`

35 = 4.905t

t = 35/4.905 = 7.135 s

The watermelon's speed at this instant can be calculated from

v = u + gt

where v is the velocity at this instant

u = initial speed of the watermelon = 0 m/s

g is the acceleration due to gravity = 9.81 m/s^2

t is the time they meet side by side = 7.135 s

substituting values, we have

v = 0 + 9.81(7.135)

v = 69.99 ≈ 70 m/s