We are to determine an equation in vertex form that models the height of the basketball above the ground versus time. We can determine this using the formula:h(t) = -16t² + vt + h₀
We are given that the basketball reaches its highest point of 12 m above the ground 0.5 s after it is released from his hands. Thus, the initial height is:h₀ = 12 mWe are also given that the ball lands on the ground after 1.3 seconds. Thus, the time it took for the ball to reach the ground is:t = 1.3 sLet's find the initial vertical velocity using the information that the basketball reaches its highest point 0.5 seconds after it is released.
The vertical velocity of the basketball at its highest point is zero since it stops before coming down.So we know:
v + (-9.8)(0.5) = 0v = 4.9 m/s
Substituting the given information into the equation above, we obtain:
h(t) = -16t² + vt + h₀h(t) = -16t² + (4.9)t + 12
The vertex form of this equation can be determined by completing the square. To complete the square, we can add and subtract the square of half of the coefficient of t from the equation above
:h(t) = -16(t² - 0.30625t) + 12
To complete the square, we add and subtract
(0.30625/2)² = 0.02368164062:h(t) = -16(t² - 0.30625t + 0.02368164062 - 0.02368164062) + 12h(t) = -16(t - 0.153125)² + 12
The vertex of this equation is the point (0.153125, 12) and is the highest point of the basketball. The coefficient of t² is negative, which means that the graph of this equation is a downward-facing equation .
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