Problem 14

Question

Distance traveled by a projectile An object is shot straight up- ward 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

The velocity after 5 seconds is 240 ft/s, and the height attained is 1600 ft.

1Step 1: Calculate the Velocity After 5 Seconds

The velocity of an object under constant acceleration can be calculated using the formula: $ v = v_0 - g \cdot t $, where $ v_0 $ is the initial velocity, $ g $ is the acceleration due to gravity, and $ t $ is the time. For this problem, $ v_0 = 400 $ ft/s, $ g = 32 $ ft/s², and $ t = 5 $ seconds. Substitute these values into the equation: $ v = 400 - 32 \cdot 5 = 400 - 160 = 240 $ ft/s. Therefore, the upper estimate for the velocity after 5 seconds is 240 ft/s.

2Step 2: Calculate the Height Attained After 5 Seconds

The height of an object under constant acceleration can be calculated using the formula: $ s = v_0 \cdot t - \frac{1}{2} g \cdot t^2 $, where $ s $ is the height, $ v_0 $ is the initial velocity, $ g $ is the acceleration due to gravity, and $ t $ is the time. Given $ v_0 = 400 $ ft/s, $ g = 32 $ ft/s², and $ t = 5 $ seconds, substitute these into the equation: $ s = 400 \cdot 5 - \frac{1}{2} \cdot 32 \cdot 5^2 = 2000 - 400 = 1600 $ ft. Thus, the lower estimate for the height attained after 5 seconds is 1600 ft.

Key Concepts

Velocity CalculationHeight EstimationConstant Acceleration

Velocity Calculation

In projectile motion, calculating the velocity of an object at a given time is crucial. This requires understanding the influence of constant acceleration on the initial launch velocity. When shooting an object upwards, gravity is working against the initial push from the surface. This constant force from gravity is denoted by the acceleration due to gravity, which is typically

32 ft/s² in situations close to Earth's surface.

Given the initial velocity, let's say 400 ft/s, use the formula:\[v = v_0 - g \cdot t\]This formula helps calculate the decreasing speed of the object as time passes due to gravity. Here,

$ v_0 $ represents the initial velocity.
$ g $ is the gravitational acceleration.
$ t $ represents the time elapsed.

For instance, after 5 seconds, the speed would decrease from 400 ft/s to 240 ft/s. Just substitute the numbers into the equation: $ v = 400 - 32 \cdot 5 = 240 $ ft/s. This accounts for the constant slowing effect gravity has on any upward-moving projectile.

Height Estimation

Estimating the height of a rising projectile allows us to understand how high an object can potentially travel before gravity begins to pull it back down. For this calculation, the following equation is essential:\[s = v_0 \cdot t - \frac{1}{2} g \cdot t^2\]This formula computes the vertical displacement, where

$ s $ is height,
$ v_0 $ is the initial launch velocity,
$ t $ represents the time elapsed,
and $ g $ is the gravitational pull against the motion at 32 ft/s².

Inserting these variables aids in finding the height at any moment during its flight path. For example, after 5 seconds, substituting the values, $ s = 400 \cdot 5 - \frac{1}{2} \cdot 32 \cdot 25 $. The calculation provides us with 1,600 ft as a lower estimate for height, considering that any effects apart from gravity are ignored.

Constant Acceleration

Understanding constant acceleration is fundamental to solving problems involving projectile motion. Acceleration is the rate at which velocity changes over time. For objects under the influence of Earth's gravity alone, this is a consistent downward force termed as gravitational acceleration. For simplicity, this is generally accepted as

32 ft/s².

In our scenario, this consistent downward acceleration is critical in determining both the velocity reduction and the trajectory shape of the projectile. The equations used in velocity and height calculations are both dependent on this consistent value. Why? Because gravity doesn't change partway through the object's motion. It exerts a constant influence, decreasing velocity and thus controlling the height the projectile can achieve. Constant acceleration simplifies our calculations and provides a uniform way to determine the effects gravity has on objects launched upwards.

Problem 14

Problem 15

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