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All courses Algebra I · F-BF.1.a 60 of 76
Build quadratic and exponential functions from context using explicit formulas, recursive processes, or calculation steps.

Write a projectile height function from initial height and initial velocity

Problem
Write the height function for a ball thrown upward from \(5~\text{feet}\) at \(48~\mathrm{ft}/\mathrm{s}\). Start with \(h(t)=-16t^{2}+vt+h_{0}\), assign each given to its term with the correct sign, and verify \(h(0)\).
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Hint

Use the projectile form \(h(t)=at^2+vt+h_0\), then match the gravity term, initial velocity, and starting height from the question.

In feet, the gravity term is \(-16t^2\), the linear term is the initial velocity, and the constant term is the starting height.

Solution walkthrough

01

Start with the vertical-motion structure

\[h(t)=-16t^2+vt+h_0\]

For height in feet and time in seconds near Earth's surface, -16 is one-half the gravitational acceleration. Its negative sign makes the trajectory bend downward.

02

Assign the initial velocity

\[v=48~\text{feet}/\text{second}~->~vt=48t\]

The ball is thrown upward, so the given initial velocity is positive 48 feet per second and supplies the linear term.

03

Assign the initial height

\[h_0=5~\text{feet}\]

The launch position is 5 feet above the reference ground level, so the constant term is positive 5.

04

Write and verify the model

\[\begin{aligned} h(t)=-16t^2+48t+5 \\ h(0)=-16(0)^2+48(0)+5=5 \end{aligned}\]

The zero-input check returns the stated initial height, confirming that every contextual value has been placed in the correct term.

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Another way

  1. Match the data to h(t)=(1/2)gt²+v₀t+h₀ with g=-32 feet per second squared.

!

Common mistake

Do not make the quadratic coefficient positive. Gravity gives the height model a negative t-squared term, while the upward initial velocity gives a positive linear term.

Solution walkthrough video