Question

1. the function y=-16t^2 +vt+8 gives the height of a platform diver above the water, in feet, t seconds after the diver leaves the platform with an initial velocity v, in feet per second. on one dive, it takes the diver 1.3 seconds to reach the surface of the water. a. what was the initial velocity, in feet per second, of the diver? b. after how much time, in seconds, since leaving the platform did the diver reach the same height as the platform? c. what was the divers maximum height, in feet, above the water? justify your answers. enter your answers and justifications on the paper.

Explanation:

Step1: Find the initial velocity \(v\) (for part a)

When the diver reaches the water surface, \(y = 0\) and \(t=1.3\). Substitute into the equation \(y=-16t^{2}+vt + 8\).

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Step2: Find the time when the diver reaches the same height as the platform (for part b)

The height of the platform is when \(t = 0\), \(y=8\). Set \(y = 8\) in the equation \(y=-16t^{2}+vt + 8\).

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We know \(t = 0\) (initial time). Solve \(-16t+14.65 = 0\), \(t=\frac{14.65}{16}=0.915625\)

Step3: Find the maximum height (for part c)

The function \(y=-16t^{2}+14.65t + 8\) is a quadratic function \(y = ax^{2}+bx + c\) with \(a=-16\), \(b = 14.65\), \(c = 8\). The time \(t\) at the vertex is \(t=-\frac{b}{2a}\)

$$ t=-\frac{14.65}{2\times(-16)}=\frac{14.65}{32}\approx0.458 $$

Substitute \(t = 0.458\) into \(y=-16t^{2}+14.65t + 8\)

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Answer:

a. The initial velocity \(v\approx14.65\) feet per second.
b. The time \(t = 0.915625\) seconds.
c. The maximum height \(y\approx11.36\) feet.