# Calculus: Antiderivatives

A paperweight is thrown straight up with an initial upward velocity of 60 meters per second and an initial height of 6 meters. The acceleration (due to gravity) of the object is given by a(t) = - 9.8 meters per second, where t is in seconds. The equation that describes the paperweight's velocity as a function of time t is:

I’m am sort of stuck on this problem, i am thinking of taking the antiderivative of the acceleration for velocity and then plus the 60 (initial upward velocity). Which turns out to be -9.8t+60+C

But I’m not feeling confident about this answer. The initial height wasn’t used as part of my solution and there should be a way to find C. I am not sure how.

Thank you so much ☺️

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1. a(t) = -9.81

v(t) = -9.81t + C
since v(0) = 60,
v(t) = -9.81t + 60

h(t) = -4.9t^2 + 60t + C
since h(0) = 6,
h(t) = -4.9t^2 + 60t + 6

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oobleck
2. Its 189.673

Because yes, like oobleck said:
a(t) = -9.81

v(t) = -9.81t + C
since v(0) = 60,
v(t) = -9.81t + 60

h(t) = -4.9t^2 + 60t + C
since h(0) = 6,
h(t) = -4.9t^2 + 60t + 6

Velocity is 0 which means -9.8t+60=0 which means t=60/9.8

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