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

Short answer

The watermelon passes Superman with velocity 35 m/sec

Step-by-step solution

Step 1. Write the given information

The watermelon is dropped from the height of 320 m

The initial velocity of watermelon is $u_w=0$
The watermelon is dropped from the distance $S=320m$

Let it hit the ground with the final velocity, $v_w$in time $t$
The watermelon is coming toward the ground with acceleration due to gravity $g=10m/s^2$
Superman heads toward the ground with velocity localid="1648187437471" $u_S=35m/sec$.

Step 2. To determine the velocity of watermelon when it passes the Superman

Since the watermelon is dropped at the instant when Superman was passing by, the distance traveled by Superman is equal to that by the watermelon. Therefore, using the equation of motion,
$u_{w}t+\frac{1}{2}g{t}^2=u_{S}t+\frac{1}{2}g{t}^2$
Substitute the known values,

$$\begin{gathered} 0\left(t\right)+\frac{1}{2}\left(10\right)t^2=35\left(t\right)+\frac{1}{2}\left(0\right)t^2 \\ 5t^2=35t \\ t=7sec \end{gathered}$$

Thus, the time at which watermelon and Superman come across is 7 sec.

Now, the velocity at which watermelon passes superman is given as,

$$\begin{gathered} v_w=u_w+gt \\ {v}_w=0+10\left(7\right) \\ {v}_w=70m/sec \end{gathered}$$

Therefore, the watermelon passes Superman with a velocity of 70 m/sec.