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

Answer 1

Answer:

Watermelon is going at a speed of $13.33\ \rm m/s$ in downward direction

Explanation:

Given

• Height of the Empire State building = 320 m
• Speed of the superman = 24 m/s

since both the motion of both  watermelon and the superman is under gravity. The acceleration of both is g in downwards direction. So the relative acceleration between them is zero and when they meet the relative distance covered by them will be the height of the Empire State building.

so let t be the time when they meet given by

$\dfrac{320}{24}\\t=13.33\ \rm s$

Now let v be the velocity of the the watermelon when they meet given by

$v=gt\\v=9.8\times13.33\\v=130.63\ \rm m/s$

Answer 2

Answer:

47.9709 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

For watermelon

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow s=\frac{1}{2}\times 9.81\times t^2\ m$

For superman

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times t+\frac{1}{2}\times 0\times t^2\\\Rightarrow s=24t\ m$

The distance travelled by the watermelon and superman will be same

$\frac{1}{2}\times 9.81\times t^2\ m=24t\\\Rightarrow t=\frac{24}{0.5\times 9.81}\\\Rightarrow t=4.89\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 4.89\\\Rightarrow v=47.9709\ m/s$

The watermelon is going at a speed of 47.9709 m/s when it passes Superman