Question

A ball is thrown from an initial height of 2 meters with an initial upward velocity of 9/ms . The ball's height h (in meters) after t seconds is given by the following. =h+2−9t5t2 Find all values of t for which the ball's height is 3 meters. Round your answer(s) to the nearest hundredth.

Answer 1

A ball is thrown from an initial height of 2 meters with an initial upward velocity of 9/ms

Balls height $h= 2 +9t -5t^2$

To find all values of t for which the ball's height is 3 meters

We plug in 3 for h and solve for t

$h= 2 +9t -5t^2$

$3 = 2 +9t -5t^2$

Solve for t

$0= -1+ 9t -5t^2$

$5t^2 - 9t + 1 = 0$

Solve using quadratic formula

$t= \frac{-b+- \sqrt{b^2-4ac} }{2a}$

$t= \frac{9+- \sqrt{(-9)^2-4*5*1} }{2*5}$

After simplifying this,

$t= \frac{9+\sqrt{61}}{10}$ = 0.11898

$t= \frac{9-\sqrt{61}}{10}$ = 1.68102

the values of t for which the ball's height is 3 meters= 0.12 sec , 1.68 sec