Question

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How high above the projection point is it after \(1.00 \mathrm{~s}\) ?

Short answer

Answer: The object's vertical displacement after 1.00 s is 23.1 m above the projection point.

Step-by-step solution

Identify the known values

Given the initial velocity (v₀) as 28.0 m/s, and the time (t) as 1.00 s, the vertical acceleration (a) will be due to gravity, which is -9.81 m/s² (negative sign indicates downward acceleration).

Choose the appropriate kinematic equation

We want to find the vertical displacement (Δy) after 1.00 s, so we should use the following kinematic equation: Δy = v₀t + 0.5at²

Substitute the known values into the equation

We have v₀ = 28.0 m/s, a = -9.81 m/s², and t = 1.00 s. Substituting these values in the equation, we get: Δy = (28.0 m/s)(1.00 s) + 0.5(-9.81 m/s²)(1.00 s)²

Calculate the vertical displacement

Now, simply calculate the resulting value of Δy: Δy = 28.0 m - 4.905 m = 23.1 m So the object is 23.1 m above the projection point after 1.00 s.

Key concepts

Vertical Displacement

Imagine an object soaring through the air after a powerful upward throw. The distance it travels from its starting point is what we call vertical displacement. In our exercise, an object is launched upwards and we're curious about how high it's gotten after just one second.

Calculating Vertical Displacement

To determine this, we turn to the kinematic equations. These are recipes that tell us the object's new position based on its initial velocity, acceleration (like gravity's tug downwards), and time in motion. The formula looks like this:
\[ \Delta y = v_0t + \frac{1}{2}at^2 \].
In practice, if you toss an object upward with a speed of \(28.0 \frac{m}{s}\), after \(1.00 s\) it covers a vertical distance. By substituting the known values, we can see that the object's height will change by \(23.1 m\), meaning that's how high it is above its starting point. Keep in mind, vertical displacement can be positive or negative depending on the object's direction—heavenward or earthward.

Initial Velocity

Think of initial velocity as the very first push that sets an object in motion. It's the starting speed, the sprinter's burst from the blocks. In kinematics, initial velocity is the speed at which an object begins its journey. In our textbook problem, the object has an initial upward velocity of \(28.0 \frac{m}{s}\).

The Role of Initial Velocity

Why does initial velocity matter? It's the springboard for our calculations—it dictates how far the object will travel before other forces, like gravity, start reining it in. A higher initial velocity means our object will climb higher before it starts to slow down and eventually reverses its course under gravity's uncompromising pull.

Acceleration Due to Gravity

Every object on Earth feels the steady pull of gravity, which inevitably drags it back to the ground. This pull, or acceleration due to gravity, is a constant that measures \(-9.81 \frac{m}{s^2}\) for objects near the Earth's surface. The minus sign you see in our exercise simply means that gravity's acceleration is directed downward, toward the center of the Earth, the opposite direction of the object's initial upward burst.

Gravity's Inevitable Influence

Gravity plays a pivotal role in kinematics problems that involve vertical motion. No matter how strong the initial push is, gravity will gradually slow down the object until it reaches the peak of its ascent. Then, it pulls the object back, increasing its speed in the downward journey. This acceleration is a crucial factor in the eventual vertical displacement of our object—it's the 'aha' moment in our story of upward motion and inevitable return.