Question

A watermelon is thrown down from a skyscraper with a speed of 7.0 m/s. It lands with an impact velocity of 20 m/s. We can ignore air resistance. What is the displacement of the watermelon?

Answer 1

A watermelon is thrown down from a skyscraper with a speed of v7.0 m/s. It lands with an impact velocity of 20 m/s. We can ignore air resistance.

What is the displacement of the watermelon?

Answer: -18

Answer 2

-17.89 m/s (or 17.89m/s downward)

edit:

if we round this to the nearest whole number, the answer is -18m/s (or 18m/s downward)

Step-by-step explanation:

recall that one of the equations of motion can be expressed as:

v² = u² + 2as,

where

v = final velocity = given as 20 m/s

u = initial velocity = given as 7.0 m/s

a = acceleration. Since it is freefalling downwards, the acceleration it would experience would be the acceleration due to gravity = -9.81 m/s²

s = displacement (we are asked to find)

simply substitute the known values into the equation:

v² = u² + 2as

20² = 7² + 2(-9.81)s

400 - 49 = -19.62s

-19.62s = 351

s = 351/-19.62

s = -17.8899

s = -17.89 m/s (or -18 m/s rounded to nearest whole number)