Problem 14

Question

An object is shot straight upward from sea level with an initial velocity of $400 \mathrm{ft} / \mathrm{sec}.$ a. Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use $g=32 \mathrm{ft} / \mathrm{sec}^{2}$ for the gravitational acceleration. b. Find a lower estimate for the height attained after 5 sec.

Step-by-Step Solution

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Answer

Velocity: 240 ft/sec. Height: 1600 ft.

1Step 1: Understand the Problem

We need to find the approximate velocity and height of an object moving upward with an initial velocity of 400 ft/sec after 5 seconds, considering the effect of gravity only.

2Step 2: Define the Velocity Equation

The velocity of an object under gravity is given by the formula: $v(t) = v_0 - g \, t$, where $v_0$ is the initial velocity, $g$ is the acceleration due to gravity, and $t$ is the time in seconds.

3Step 3: Calculate Velocity After 5 Seconds

Plug in the values: $v_0 = 400 \, ext{ft/sec}$, $g = 32 \, ext{ft/sec}^2$, and $t = 5 \, ext{sec}$ into the equation: $v(5) = 400 - 32 \times 5$. Simplifying gives $v(5) = 400 - 160 = 240 \, ext{ft/sec}$. This is the upper estimate for the velocity after 5 seconds.

4Step 4: Define the Height Equation

The height of an object under constant acceleration can be described by: $s(t) = v_0 t - \frac{1}{2} g t^2$, where $s(t)$ is the height at time $t$.

5Step 5: Calculate Height After 5 Seconds

Plug in $v_0 = 400 \, ext{ft/sec}$, $g = 32 \, ext{ft/sec}^2$, and $t = 5 \, ext{sec}$ into the height equation: $s(5) = 400 \times 5 - \frac{1}{2} \times 32 \times (5)^2$. Simplify to get $s(5) = 2000 - 400 = 1600 \, ext{ft}$. This is the lower estimate for the height attained after 5 seconds.

Key Concepts

VelocityAcceleration Due to GravityProjectile Motion

Velocity

Velocity is a key concept in kinematics that describes how fast an object changes its position over time. In simpler terms, it provides the speed and direction of a moving object. In our problem, an object is initially shot upwards with a velocity of 400 ft/sec. This means that in the absence of any forces, this object would cover 400 feet in one second along the upward direction. However, due to gravity, this velocity decreases over time.

Initial velocity ($v_0$) is given as 400 ft/sec.
Gravity ($g$) pulls the object downward, decreasing its upward velocity.
At any time ($t$), the velocity ($v(t)$) can be calculated using the formula: $v(t) = v_0 - g \, t$.

In the given exercise, using the formula, we calculated that after 5 seconds, the velocity of the object decreases to 240 ft/sec. This decrease is directly due to the constant pull of gravity acting on the object. Understanding velocity and its calculation is critical for determining how an object will move through space.

Acceleration Due to Gravity

Gravity is a fundamental force that pulls objects toward one another, with the Earth's gravity being the most commonly experienced form. The acceleration due to this force is constant near the Earth's surface and affects everything that moves, including objects in projectile motion.

The constant acceleration due to gravity ($g$) is approximately 32 ft/sec² on Earth.
This means that every second, an object's velocity changes by 32 ft/sec toward Earth.
In projectile motion, when an object is moving upward, $g$ reduces its upward velocity until it eventually reaches the peak height and starts coming back down.

In the given exercise, gravity acts as the sole force that counteracts the upward motion. It causes a reduction in velocity and impacts how high the object can go. It is this reliance on $g$ that makes calculations and predictions in physics both reliable and insightful.

Projectile Motion

Projectile motion refers to the movement of an object that is projected into the air and moves under the influence of gravity alone. It is characterized by an initial force that propels the object and the gravitational force that planes it back down. Here's how this applies to our exercise:

The initial force in projectile motion is the object's launch velocity. In this case, it is 400 ft/sec directly upwards.
Gravity is the only force acting on the object after the initial launch, pulling it downward at 32 ft/sec².
To calculate the object's height after a specific time, we use the formula: $s(t) = v_0 t - \frac{1}{2} g t^2$.

In the provided example, we found that the object reaches an approximate height of 1600 feet after 5 seconds. This height reflects how projectile motion is influenced by both the initial velocity and the opposing pull of gravity. Understanding projectile motion is essential for predicting how objects behave when thrown into the air, whether it be a basketball shot, a rocket launch, or even a simple toss of a ball.

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