Solving a word problem using a quadratic equation with irration

Question

A ball is thrown from an initial height of 2 meters with an initial upward velocity of 13/ms  The ball's height h (in meters) after t seconds is given by the following. =h+2−13t5t2 Find all values of t for which the ball's height is 9 meters.

Answer 1

All the values of time t for which h(t) = 9 satisfy

-5t² + 13t + 2 = 9

That gives the quadratic equation in standard form

-5t² + 13t - 7 = 0

The solutions are of the form

t = (-13 ± √(13² - 4×(-5)×(-7)))/2×(-5)

That simplifies to two solutions

t = (13 + √29)/10 = 1.8385

t = (13 - √29)/10 = 0.7619

Answer 2

the standard formula for height in meters is h(t) =-4.9t^2 +vot + ho where t=time in seconds, vo= initial velocity and ho=initial height.

9 = -4.9t^2 + 13t +2

-4.9t^2 + 13t -7 = 0

4.9t^2 -13t + 7 = 0

t= 13/9.8 + or - (1/9.8)sqr(169-4(4.9)(7))

= 1.33 + or - (1/9.8)sqr(31.8) = 1.33 + or - 5.64/9.8 = 1.33 + or - .575 = 1.905 or 0.755 seconds

but you have a different equation, which is close, seemingly 5t^2 -13t +2. 4.9 is close to 5, but the signs are wrong. the t^2 term is for deceleration due to gravity and is negative. the 13t term is for velocity and should be positive if thrown up.