A ball is thrown straight up into the air from the top of a building standing at 50 feet with an initial velocity of 65 feet per second the height of the ball in feet can be modeled by the following function: h(t)=-16t^2++16t+96 When does the ball reach its maximum height?

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Question:

A ball is thrown straight up into the air from the top of a building standing at 50 feet with an initial velocity of 65 feet per second the height of the ball in feet can be modeled by the following function: h(t)=-16t^2++16t+96 When does the ball reach its maximum height?

Answers

Answer:0.5 seconds (at 100 feet in the air). Step-by-step explanation:So, the height of the ball can be modeled by the function: [tex]h(t)=-16t^2+16t+96[/tex]Where h(t) represents the height in feet after t seconds. And we want to find its maximum height. Notice that our function is a quadratic. Therefore, the maximum height will occur at the vertex of our function. The vertex of a quadratic function in standard form is given by the formula: [tex](-\frac{b}{2a}, f(-\frac{b}{2a}))[/tex]In our function, a=-16; b=16; and c=96. Find the x-coordinate of the vertex: [tex]x=-\frac{(16)}{2(-16)}=1/2[/tex]So, the ball reaches its maximum height after 0.5 seconds of its projection. Notes: To find it’s maximum height, we can substitute 1/2 for our function and evaluate. So: [tex]h(1/2)=-16(1/2)^2+16(1/2)+96[/tex]Evaluate: [tex]h(1/2)=-4+8+96=100[/tex]So, the ball reaches its maximum height of 100 feet 0.5 seconds after its projection.

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