Question

a ball is thrown from an initial height of 6 feet with an initial upward velocity of 37 ft/s. the balls height h (in feet) after t seconds is given by the following.

$h = 6 + 37t - 16t^{2}$

find all values of t for which the balls height is 26 feet.

round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Substitute \( h = 26 \) into the equation

Given \( h=6 + 37t-16t^{2} \), when \( h = 26 \), we have \( 26=6 + 37t-16t^{2} \).
Rearrange it to the standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 16t^{2}-37t + 20=0 \). Here \( a = 16 \), \( b=-37 \), \( c = 20 \).

Step2: Use the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)

First, calculate the discriminant \( \Delta=b^{2}-4ac=(-37)^{2}-4\times16\times20 \)
\( =1369 - 1280=89 \).
Then \( t=\frac{37\pm\sqrt{89}}{2\times16}=\frac{37\pm9.43}{32} \).
For the plus - case: \( t=\frac{37 + 9.43}{32}=\frac{46.43}{32}\approx1.45 \).
For the minus - case: \( t=\frac{37-9.43}{32}=\frac{27.57}{32}\approx0.86 \).

Answer:

\( t = 0.86\) or \(t = 1.45\)