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A student drops a water balloon out of a dorm window 17 m above the ground. What is its speed when it hits the ground?

User Sostom
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1 Answer

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11 votes
so the answer is about 18.24 m/s

this is why:

the equation for calculating the height position from the acceleration due to gravity is
h(t) = -1/2(g)t^2 + vt + p

so in this case g, or the acceleration of gravity is 9.81, the initial velocity (v) is 0, and the initial position (p) is 17. once we plug in these numbers we get the equation h(t) = -4.91t^2 + 17

we know that height is zero when it hits the ground so we can use the quadratic equation to solve for when y is zero. since time cant be negative, then out answer is the positive zero of this equation, or 1.86.

once we know this then we can plug this amount of time into the equation
v(t) = gt + v

this means that the velocity after a certain time (t) is equal to the gravitational force, in this case 9.81, multiplied by the amount of time passed plus the initial velocity (v).
since we know the initial velocity is zero and the time that the balloon hit the ground (1.86), we can multiply the gravitational force, 9.81, by the time when it hits the ground (1.86) to get 18.24 m/s
User Khurram Ishaque
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