Calculus Velocity and Acceleration

Question

Here's the question:

I know that:

$a(t) = -9.8$

So I integrated the acceleration function to find the velocity:

$v(t) = -9.8t + c$


And because $v(0) = -5$, I can determine that $c = -5$, thus:

$v(t) = -9.8t - 5$


I then integrated the velocity function to find the position:

$s(t) = -4.9t^2 - 5t + c'$

And because at time $t= 0$, the position $s(t=0) = 0$, $c' = 0$, thus:

$s(t) = -4.9t^2 - 5t$

Then, when the ball hits the ground, $s(t) = 30$, so:

$4.9t^2 + 5t + 30 = 0$

This is my working so far and I just wanted to check if I did it right since I was a bit confused by the question. To find $t$ now would I just sub in to the quadratic formula? Any input would be appreciated, thanks!

Answer 1

Almost all of it is correct, only the last equation is wrong:

The ground is 30 meters UNDER the bridge, so you want the time at which the value od $s(t)$ is equal to $-30$.

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It will be very helpful for you to understand why I was able to very quickly notice that something is wrong with your solution. Here is the thought train that got me there:

• OK, looks pretty much OK, the OP found $s(t)$ to be some sort of quadratic expression in $t$.
• I see, now he's solving the equation $s(t)=30$
• But wait, that means he's solving an equation $c_1 t^2 + c_2 t + 30 = 0$! But that's no good! Because $c_1$ and $c_2$ are both positive, $c_1 t^2 + c_2 t + 30$ will be positive if the time is positive!
  • This is not possible: from common sense, I know that the ball must hit the ground. There must have been a sign mixup somewhere.