Question

You are standing on top of the Empire State Building (height 330 [m]) when Superman flies past. He is headed straight down with a steady speed of 35 [m/s]. (Superman can do things like this.) At the instant he goes past, you drop a 10 [kg] lead ball over the edge. At what height above the sidewalk does the ball pass Superman?

Answer 1

Answer:

The ball will be overtake superman after travelled distance 248.5 m.

Explanation:

Given that,

Speed of superman = 35 m/s

Height = 330 m

Mass of ball = 10 kg

Let the height at which they are at same position be S.

For superman,

We need to calculate the time

Using equation of motion

$S=ut$...(I)

For ball,

$S=u't+\dfrac{1}{2}gt^2$

From equation (I) and (II)

$ut=u't+\dfrac{1}{2}gt^2$

$ut=0+\dfrac{1}{2}\times g\times t^2$

$u=\dfrac{1}{2}\times g\times t$

$t=\dfrac{2u}{g}$

Put the value into the formula

$t=\dfrac{2\times35}{9.8}$

$t=7.1\ sec$

We need to calculate the distance travelled

Using formula of distance

$d=v\times t$

Put the value into the formula

$d=35\times 7.1$

$d=248.5\ m$

Hence, The ball will be overtake superman after travelled distance 248.5 m.