Question

A ball is thrown from an initial height of 3 ft with an initial upward velocity of 27ft/s The balls height h (in feet) after t seconds is given by the following. h=3+27t-16t^2. Find all values of t for which the balls height is 13 ft. ...?

Answer 1

we are looking for values of t for which
h=13 -16t^2+27t+3=13
or
 -16t^2+27t-10=0
 
This is a quadratic equation with two solutions: t1=(27 - sqrt(89))/32
 t2=(27 + sqrt(89))/32

Hope this helps

Answer 2

Answer:

t = 0.55 s and t = 1.14 s

Explanation:

as we know that height in terms of time is given as

$h = 3 + 27t - 16 t^2$

now we have to find the time at which its height is h = 13 ft

so here we can say

$13 = 3 + 27 t - 16 t^2$

$16 t^2 - 27 t + 10 = 0$

$t = \frac{27 \pm \sqrt{27^2 - 4(16)(10)}}{2(16)}$

$t = \frac{27 \pm 9.43}{32}$

$t = 1.14, 0.55$

so it will be at h = 13 ft height at t = 0.55 s and t = 1.14 s