a watermelon is thrown down from the 7th floor of urey hall at ucsd. if the initial height of the watermelon is 21.0 meters, and it is thrown straight downward with an initial downward velocity of 3.00 m/s. how far will the watermelon have fallen from its starting height after 1.50 seconds?

The watermelon has fallen 15.53625m

Step-by-step explanation:

x=vi*t+.5*a*t^2

x=3*1.5+.5*9.81*1.5^2

x=15.53625m

The answer would be 15.5

1) on the number line, -1.5 can be seen as going backwards 1.5.
there are -4 groups of -1.5
this means that you have to move in the opposite direction that you would normally move 1.5
4 groups of 1.5 is 6 (because you have to move in the positive direction)
6 is the answer

2) 975 feet below the surface