We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# help please

+1
134
3
+80

A ball travels on a parabolic path in which the height (in feet) is given by the expression -16t^2+64t+31, where t is the time after launch. What is the maximum height of the ball, in feet?

Dec 27, 2018

### 3+0 Answers

#1
+5057
+3

$$\text{can you use calculus?}\\ \text{If so we can find the first derivative of the height (it's vertical velocity)}\\ \text{and solve the the }t \text{ that makes this zero}\\ \text{(the ball stops for a moment at the top of it's arc)}\\ v_z(t) = \dfrac{d}{dt} h(t) = -32t + 64\\ v_z(t) = 0 \Rightarrow t = 2\\ h(2) = -16(4) + 64(2) + 31 = 95$$

$$\text{If you can't use calculus (and I suspect you can't)}\\ \text{we can rewrite }h(t) \text{ in standard parabola form}\\ -16t^2 + 64t + 31 = \\ -16(t^2 -4t) + 31 = \\ -16((t-2)^2-4)+31 = \\ -16(t-2)^2 + 64 + 31 = \\ -16(t-2)^2 + 95\\ \text{and it should be clear this reaches a maximum of }\\ h=95 \text { at }t=2$$

.
Dec 27, 2018
#2
+18311
+2

Max height will be at time, t   =   -a/(2b) =   - 64/(2*-16) = 2 seconds

Substitue this time into the equation to find max height

-16(2^2) + 64(2)+31 = 95 ft max

Dec 27, 2018
#3
+647
0

Great method, EP!

CoolStuffYT  Dec 27, 2018