HELP ME SOON PLZ

Question

A ball is thrown from an initial height of 5

 feet with an initial upward velocity of 23/fts

. The ball's height h

 (in feet) after t

 seconds is given by the following.

=h+5−23t16t2

Find all values of t

 for which the ball's height is 13

 feet.Round your answer(s) to the nearest hundredth.

Answer 1

You probably meant h(t) = -16t^2 + 23t + 5 as -23t would mean throwing down, not up, 5 is the initial height and =16t^2 is the effect of gravity, as it decelerates at 32 feet per second per second. Also maybe you also meant 24t as it simplifies the problem that way.

but for 23t

for height of 13 set h= 13 and solve for t. There should be two solutions as it reaches 13 on the way up, and then on the way back down

plug in some simple numbers for t and solve for h

at zero seconds, it's at 5 feet

at 1 second it's at 12 feet

at 2 seconds it's at 3 feet

at 1 1/2 seconds it's at 3 1/2 feet

at 1 1/4 seconds it's at 9 3/4 feet

at 1 1/8 seconds it's at 10 5/8 feet

at 23/32 = 0.72 seconds it's at 13.27 feet the maximum height

It reaches 13 feet, slightly before and slightly after 0,72 seconds

-16t^2 + 23t +5 = 13

-16t^2 +23t - 8 = 0

use the quadratic formula

t = 23/32 + or - (1/32)sqr(23^2-64(8)) =

= 23/32 + or - (1/32) sqr(529-512) = 23/32 + or - (1/32)sqr(17) = (23+ or -sqr17)/32 =

= 0.85 or 0,59 seconds at 13 feet

max at 0.72

the 13 feet height is 0.13 + 0,72 or 0.72-0.13

It reaches maximum height at t = 23/32 = 0.72 seconds

h'(t)= -32t +23=0, t = 23/32

had the problem been

-16t^2 + 24t - 8 = 0

then divide by 8

-2t^2 +3t -1 = 0

2t^2 -3t +1 = 0

t^2 -3t/2 =-1/2

(t^2 -3/4)^2 = -1/2 +9/16

t -3/4 = + or - sqr(1/16)= + or - 1/4

t =3/4 +1/4 = 1 or 0.5 seconds to reach 13 feet

max height at t=24/32 = 3/4 = 0.75 seconds

It comes out a little more evenly with -24t instead of -23t for the velocity.

sometimes we hit the 3 key, when we intend the 4 key. It happens.