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Difficulty: MediumQuadratic Equations

A projectile is launched from the ground. Its height, in feet, tt seconds after launch is modeled by the function h(t)=−16t2+v0th(t) = -16t^2 + v_0 t, where v0v_0 is the initial upward velocity in feet per second. If the projectile reaches its maximum height of 144144 feet, what is the value of v0v_0?

Answer: 96

Answer

The correct answer is 96. The initial velocity of the projectile must be 96 feet per second to reach a maximum height of 144 feet.
The maximum height of a projectile modeled by a quadratic function is the yy-value of its vertex. For h(t)=−16t2+v0th(t) = -16t^2 + v_0 t, the time tt at the vertex is given by t=−b2a=−v02(−16)=v032t = -\frac{b}{2a} = -\frac{v_0}{2(-16)} = \frac{v_0}{32}. Substituting this value of tt back into the height equation yields the maximum height: h(v032)=−16(v032)2+v0(v032)=−16(v021024)+v0232=−v0264+v0232=v0264h\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0^2}{1024}\right) + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{v_0^2}{64}. Given that the maximum height is 144144 feet, we set v0264=144\frac{v_0^2}{64} = 144, which simplifies to v02=144×64v_0^2 = 144 \times 64. Taking the square root of both sides gives v0=12×8=96v_0 = 12 \times 8 = 96.

Step-by-Step Solution

1
Find the time tt at which the maximum height occurs using the vertex formula.
t=v032t = \frac{v_0}{32}
For a quadratic function of the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the vertex occurs at x=−b2ax = -\frac{b}{2a}. Here, a=−16a = -16 and b=v0b = v_0, so the vertex is at t=−v02(−16)=v032t = -\frac{v_0}{2(-16)} = \frac{v_0}{32}.
2
Substitute the time t=v032t = \frac{v_0}{32} into the height function to express the maximum height in terms of v0v_0.
h(v032)=v0264h\left(\frac{v_0}{32}\right) = \frac{v_0^2}{64}
Evaluating h(v032)h\left(\frac{v_0}{32}\right) gives −16(v032)2+v0(v032)=−16(v021024)+v0232=−v0264+v0232=v0264-16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0^2}{1024}\right) + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{v_0^2}{64}.
3
Set the maximum height expression equal to the given maximum height of 144144 feet and solve for v0v_0.
v0=96v_0 = 96
Setting v0264=144\frac{v_0^2}{64} = 144 yields v02=144×64=9216v_0^2 = 144 \times 64 = 9216. Taking the positive square root (since v0v_0 represents a positive initial upward velocity) gives v0=144×64=12×8=96v_0 = \sqrt{144} \times \sqrt{64} = 12 \times 8 = 96.

Key Concept

Determining the vertex coordinates of a quadratic function to find maximum value in context.
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Topics

Quadratic Equations

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