Question

Ball is kicked from a height of 24 feet at an initial velocity of 16 ft/sec. Approximately how long does it take before the ball lands on the ground?

Answer 1

So far ive gotten to 0=4t^2 + 4t + but i dunno if it can factor out. Did I screw up somewhere?

Answer 2

Unspecified is the direction in which the ball is kicked, but that is vital to answering the question...

Answer 3

What??

Answer 4

Without knowing the direction the ball gets that velocity of 16 ft/s, it is impossible to solve the problem.

Answer 5

If you use the Projectile Motion Equation then it is just a matter of plugging in numbers

Answer 6

The answer is very different if the ball goes straight up, or if it is kicked horizontally, or down toward the ground.

No, it's not, unless you know the direction.

Answer 7

Ok, then lets just say it is kicked upwards, now what?

Answer 8

h(t) = -1/2gt^2 + vt + h

Answer 9

Ball going straight up has equation of motion
\[h(t) = -16t^2 + 16t +24\]

Ball kicked horizontally from a 24 foot high platform would have equation of motion
\[h(t) = -16t^2 + 0t + 24\]

Okay, we'll do straight up. You're looking to find the time at which it hits the ground, or \(h(t) = 0\), so
\[-16t^2 + 16t + 24 = 0\]

Answer 10

Factor out a 4:
\[-4t^2 + 4t + 6 = 0\]

Answer 11

Ok, I just made a mistake in my math, I did same thing but I got -3t^2 + 4t + 8

I just messed up the 24/4

Answer 12

I'd probably just do the quadratic formula at this point, with \(a = -4, b = 4, c = 6\)

\[t = \frac{ -b \pm \sqrt{b^2 -4ac}}{2a}\]

Answer 13

one of the values will be negative, and we can ignore that one.

Answer 14

So t= 84.667?

Answer 15

no...want to show me your work? that's off by about a factor of 45 or so

Answer 16

that'd be amazing hang time, ball would look like it just hung there on a string or something :-)

Answer 17

Calculator is really not working today :( After recalculating I got 13.166

Answer 18

write out your numbers for me

Answer 19

what does the equation look like before you punch it into the calculator?

Answer 20

Cant write it out because of square root symbol. Hold on one sec, I will try it again

Answer 21

just say sqrt(stuff) where you want the square root

-b +/- sqrt(b^2 - 4ac) etc

Answer 22

10.5? my last step was -84/-8

Answer 23

no. I have to go now, so I'll just show you how I do it:

\[a = -4, b = 4, c = 6\]\[t = \frac{ -b \pm \sqrt{b^2 -4ac}}{2a}\]
\[t = \frac{-4 \pm\sqrt{(4)^2 - 4(-4)(6)}}{2(-4)} = \frac{-4\pm\sqrt{16+96}}{-8}\]
you should be able to finish from there

Answer 24

15

Answer 25

lolol omg I hate math sooooo much thx for your help

Answer 26

no, how do you get that?
16+96 = 112. 112 = 7*16 so sqrt(112) = sqrt(7)*sqrt(16) = 4sqrt(7)

answer is
\[\frac{-4\pm4\sqrt{7}}{-8}\]

Answer 27

Ohhh yea the sgrt. I forgot about it

Answer 28

ok, that makes a lot more sense

Answer 29

Would that also be the maximum height that the ball can reach?

Answer 30

No, the maximum height will be at the vertex of the parabola. If you have a parabola written in the form \[y = ax^2 + bx + c\]the vertex x-coordinate is given by \[x = -\frac{b}{2a}\]and the vertex y-coordinate can be found by plugging that value of \(x\) back into the equation.

Also, the vertex is exactly at the midpoint on the x-axis between the two solutions where the parabola = 0. Note that this is NOT necessarily half of the flight time! If you start at a different elevation than you land, the vertex will be before the halfway mark in the flight.