A ball is thrown into the air with an upward velocity of 44 ft/s. Its height \( h \) in feet after \( t \) seconds is given by the function [tex]h = -16t^2 + 44t + 9[/tex]. What is the height of the ball after 2 seconds?

h = − 16 t 2 + 44 t + 9 s u b s t i t u t e t = 2 : h = − 16 ⋅ 2 2 + 44 ⋅ 2 + 9 = − 16 ⋅ 4 + 88 + 9 = − 64 + 97 = 33 A n s w er : 33 f ee t s .

Answered by Anonymous | 2024-06-10

t h e f u n c t i o n : h ( t ) = − 16 t 2 + 44 t + 9 h = 2 ⇒ h ( 2 ) = − 16 ⋅ 2 2 + 44 ⋅ 2 + 9 = − 16 ⋅ 4 + 88 + 9 = 33 A n s . t h e h e i g h t o f t h e ba ll a f t er 2 seco n d s i s 33 f ee t s

Answered by kate200468 | 2024-06-10

After substituting 2 seconds into the height function, we find that the height of the ball is 33 feet after 2 seconds. This is calculated by evaluating the quadratic equation given. The final result indicates the position of the ball at that specific moment in time.
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Answered by Anonymous | 2024-12-24