Answer:
A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that balls height h (in feet) after t seconds is give by:
<u>h= 5 + 29t -16t^2</u>
Explanation:
h= 5 + 29t -16t^2
a time when the ball's height will be 17 ft
17 = 5 + 29t -16t2
0 = -17 + 5 + 29t -16t2
0 = -12 + 29t - 16t2
Using the quadratic equation:
t = (-29±√(292-(4*(-16)*(-12))))÷2(-16)
= (-29±√(841 - 768))÷(-32)
= (-29±√(73))÷(-32)
= (-29 + 8.544)÷(-32) or (-29 - 8.544)÷(-32)
= (-20.456)÷(-32) or -37.544÷(-32)
= 0.64 or 1.17
So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.
It's very hard to see the self-portrait, so I can't identify him.
