Position, velocity, and acceleration question

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An object is $100$ feet above ground and has an initial velocity of $15$ m/s upwards and it is being accelerated by gravity. What is the equation for the height?

So right away, I know that it would be $p(t)=100$ when $t=0$ and $v(t)=15$ and $a(t)=-9.8t$. Since velocity is the integral of acceleration, then $v(t)=15-\int_0^t 9.8 dt = 15-9.8t$. Then the position is the integral of velocity and so $p(t)=100+\int v dt = 100+ \int_0^t15-9.8t dt=100+15t-4.9t^2$

Is that correct?

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Yes, that is correct.

You could also use the formula that is derived by your logic, that $p(t)=\frac{1}{2}at^2+tv(0)+p(0)$.

In this case $a=-g$ because gravity.