# Height and velocity of ball thrown vertically

A ball is thrown upward from roof of 32 foot building with velocity of $112$ ft/sec. The height after $t$ seconds is: $s(t)=32+112t-16t^2$.
(a) Find the maximum height that the ball reaches. (answer: $228$)
(b) Find the velocity of the ball when it hits the ground. (answer: $-120.79735$)

(a) I took the derivative of the height function to get the velocity function and set it equal to zero, since the maximum height will be at the top of the inverted parabola, and at point the velocity (derivative) is zero (right?):

$v(t)=112-16t=0 \implies t=7$, then I substituted this $t$ into $s(t)$ to get $32$, which is wrong.

(b) the velocity of the ball when it hits the ground is when height=$0$, right? I set $s(t)=30$? How do I solve this?

Thank you.

You made a minor mistake: $v(t)$ should be $112 - 32t$. This should make everything correct. As for the second part, ground means $s(t)=0$. So find the corresponding time and plug into $v(t)$ to find the impact velocity.
• Wouldn't it be $112$ since the derivative of $112t$ is just $112$? can you show some more steps please? Thanks +1 Commented Oct 3, 2014 at 2:52
Given $s(t) = 32 + 112 t - 16 t^{2}$ then $v(t)$, being the derivative of $s(t)$, is $v(t) = 112 - 32 t$. If $v(t) = 0$ then $t = 7/2$. Now, $s(7/2) = 32 + 56 \cdot 7 - 4 \cdot 49 = 228$.
To find the velocity at impact solve for $s(t) = 0$. This yields $16 t^{2} - 112 t - 32 = 0$, or $t^{2} - 7 t - 2 = 0$. Solving this equation leads to the two possible values \begin{align} t = \frac{7}{2} \pm \frac{\sqrt{49+8}}{2} = \frac{7}{2} \pm \frac{\sqrt{57}}{2}. \end{align} The negative component is negative and is to be tossed out of consideration leaving $2 t_{I} = 7 + \sqrt{57}$. Now, \begin{align} v(t_{I}) = 112 - 32 t_{I} = 112 - 16 \cdot 7 - 16 \cdot \sqrt{57} = - 16 \sqrt{57} = -120.79735\cdots . \end{align}
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