Vertical projectile motion with initial height and velocity

Expert intro: This problem uses the standard constant-acceleration model for vertical motion near Earth’s surface. The key is to track position, velocity, and acceleration as functions of time using the given initial height and initial velocity.

Detailed walkthrough

Identify the motion model

For vertical projectile motion near Earth’s surface, acceleration is constant and directed downward. In feet and seconds, the usual model is

$s(t)=s_0+v_0t-16t^2,$

where $s_0$ is the initial height and $v_0$ is the initial velocity. The negative $16t^2$ term comes from gravity, since the acceleration is $-32\text{ ft/s}^2$.

Here, the object starts at height $129$ feet with initial velocity $87$ ft/s upward, so $s_0=129$ and $v_0=87$.

Write the position, velocity, and acceleration functions

Substitute the given values into the position formula:

$s(t)=129+87t-16t^2.$

Differentiate position to get velocity:

$v(t)=s'(t)=87-32t.$

Differentiate again to get acceleration:

$a(t)=v'(t)=-32.$

So the three functions are

$\boxed{s(t)=129+87t-16t^2,\quad v(t)=87-32t,\quad a(t)=-32}.$

Why these formulas work

The position function includes all three pieces of information the problem gives: initial height, initial velocity, and gravity. The velocity function is linear because acceleration is constant. The acceleration function is constant because the problem describes ideal projectile motion with no air resistance.

A quick check confirms the model is consistent: at $t=0$, the height is $129$ feet and the velocity is $87$ ft/s, exactly as stated.

Common mistake to avoid

Do not use $-32t^2$ for acceleration. The $-32$ is the acceleration, while $-16t^2$ appears in the position function after integrating acceleration twice. Mixing these up is one of the most common errors in projectile-motion questions.

💡 Pitfall guide

A frequent mistake is writing the acceleration as $-16$ because students remember the position formula $s(t)=s_0+v_0t-16t^2$. That coefficient belongs to the height function, not the acceleration. Another common issue is forgetting that the initial velocity is positive when the object is launched upward, so the velocity starts at $87$ rather than $-87$. If the problem later asks for when the object reaches the ground or its maximum height, those follow from the same functions, but the first step is always to keep the three functions separate and consistent.

🔄 Real-world variant

If the launch were downward instead of upward, the only immediate change would be the sign of the initial velocity. For example, if the same object were thrown downward from $129$ feet with speed $87$ ft/s, the position function would become $s(t)=129-87t-16t^2$, while the acceleration would still be $a(t)=-32$. If the problem asked for metric units instead of feet, the constant acceleration would change to approximately $-9.8\text{ m/s}^2$, and the coefficient in the position formula would change accordingly.

🔍 Related terms

constant acceleration model, vertical motion equation, position velocity acceleration