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 diver's maximum height, in feet, above the water?

Answer 1

The initial velocity v for the diver is 16 feet per second. To find the time at the same height as the platform and the maximum height, we use the vertex of the quadratic equation and the given formula.

Step-by-step explanation:

The function given is h(t) = -16t^2 + vt, where h(t) is the height above the water at time t, and v is the initial velocity. We are told it takes 1.42 seconds for the diver to hit the water, leading to h(1.42) = 0. Solving for the initial velocity v, we set the equation to 0 and solve for v:

• 0 = -16(1.42)^2 + v(1.42)
• v = 16(1.42) / 1.42
• v = 16 feet per second

To find the time when the diver reaches the same height as the platform, we solve for t when h(t) equals zero. There will be two solutions for t in the quadratic equation: one at t=0 (when the diver just leaves the platform) and the other at the time when the diver returns to the platform height before hitting the water.

To find the maximum height, we look for the vertex of the parabola represented by the equation, which is at t = -b/2a. This gives:

• t = -v / (2*(-16))
• t = v / 32
• Maximum height: h(t) at this t value.

By substituting v = 16 into the equation for time, we get t and then can find the maximum height by placing this t back into h(t).