A student throws a water balloon vertically downward from the top of a building.?

Question

The balloon leaves the thrower's hand with a speed of 6.30 m/s. Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand.

What is its speed after falling for a time 1.85 s?
How far does it fall in a time of 1.85 s?
What is the magnitude of its velocity after falling a distance 11.3 m?

Answer 1

Motion will follow
y(t)=-6.30*t-.5*g*t^2

and
v(t)=-6.3-g*t

for the first two, plug in 1.85 seconds and solve for y and also v - take the magnitude since they will be negative

to solve the third, plug 11.3 into y and solve for t taking the positive root. Then plug that t into v(t).

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