Question

the function ( h(t)=-16 t^{2}+v t + 10 ) 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 1.42 seconds to reach the surface of the water. - what was the initial velocity, ( v ), in feet per second, of the diver? - after how much time, in seconds, since leaving the platform did the diver reach the same height as the platform? - what was the divers maximum height, in feet, above the water?

Explanation:

Step1: Find the initial velocity \(v\)

When the diver reaches the water, \(h(t) = 0\) and \(t = 1.42\). Substitute into the equation \(h(t)=-16t^{2}+vt + 10\):

$$
LATEXBLOCK0
$$

Step2: Find the time when the diver reaches the same height as the platform

The height of the platform is \(h = 10\). Set \(h(t)=10\), so \(-16t^{2}+vt + 10=10\). Substitute \(v = 15.68\):

$$
LATEXBLOCK1
$$

We have \(t = 0\) (initial time) or \(-16t+15.68 = 0\). Solving \(-16t+15.68 = 0\) gives \(t=\frac{15.68}{16}=0.98\)

Step3: Find the maximum height

The function \(h(t)=-16t^{2}+15.68t + 10\) is a quadratic function \(y = ax^{2}+bx + c\) with \(a=-16\), \(b = 15.68\), \(c = 10\). The \(t\) - value of the vertex is \(t=-\frac{b}{2a}=-\frac{15.68}{2\times(-16)}=\frac{15.68}{32}=0.49\)
Substitute \(t = 0.49\) into \(h(t)\):

$$
LATEXBLOCK2
$$

Answer:

• The initial velocity \(v\) is approximately \(15.68\) feet per second.
• The diver reaches the same height as the platform after \(0.98\) seconds.
• The maximum height is approximately \(13.84\) feet.