Question

A ball is thrown from an initial height of 4 feet with an initial upward velocity of 17ft/s. The ball's height h (in feet) after t seconds is given by the following h=4+17t-16tsquare. Find all values of for which the ball's height is 8 feet

Answer 1

set equation equal to 8: 8 = 4 + 17t - 16t^2, use quadratic formula to solve for t. You should receive two real solutions.

Answer 2

So far I got -4+17t-16t^2=0
but I don't know how to solve it. Can you help me with that?

Answer 3

sure! np

Answer 4

\[t = -4 +17t - 16t^2 \]
using quadratic formula, where a = the coefficient of t^2, b = the coefficient of t, and c = the constant term. we obtain the following two solutions of t. \[t = \frac{ -17 + \sqrt{17^2 - 4(-16)(-4)} }{2(-16) }\]
and \[\frac{ -17 - \sqrt{17^2 - 4(-16)(-4)} }{2(-16) }\]
solving the first equation we have\[\frac{ -17 + \sqrt{33} }{-32 }\]
the evaluation of the second equation is identical save for the sign change:\[\frac{ -17 - \sqrt{33} }{-32 }\] they both give you real positive values of t for which the height of the ball is 8 feet

Answer 5

so at t = .711 seconds and at t = .352 seconds the ball is at 8 feet

Answer 6

plug these in to make sure (I already did)

Answer 7

thanks a lot

Answer 8

np!

Answer 9

i have to solve bunch of them, can you solve one more for me so I will get the technice. Because I tried to do on my own but my answers didn't come up wright

Answer 10

A ball is thrown from an initial height of 4 feet with an initial upward velocity of 31ft/s . The ball's height h (in feet) after t seconds is given by the following.

h=4+31t-16tsquare

Find all values of for which the ball's height is 18 feet.