91Ó°ÊÓ

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" $u_S=35m/sec$.

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.