Problem 28 - Calculus III

Question

A projectile is fired at a height of 1.5 m above the ground with an initial velocity of 100 m/sec and at an angle of 30° above the horizontal. Use this information to answer the following questions: Determine the range of the projectile.

Step-by-Step Solution

Answer

The range of the projectile is approximately 883 meters.

Step 1: Breakdown of Initial Velocity

The initial velocity of the projectile is given as 100 m/sec at an angle of 30°. We should first find the horizontal and vertical components of the initial velocity using trigonometric functions:\[v_{0x} = 100 \times \cos(30°) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \text{ m/s}\v_{0y} = 100 \times \sin(30°) = 100 \times \frac{1}{2} = 50 \text{ m/s}\]

Step 2: Calculate Time of Flight

Next, find the time of flight. To do this, we'll first find the time taken to reach the maximum height where vertical velocity becomes zero. We use the formula:\[v = v_{0y} - gt\]At maximum height, \(v = 0\):\[0 = 50 - 9.8t\]Solving for \(t\):\[t = \frac{50}{9.8} \approx 5.10 \text{ seconds}\]Since the projectile is initially 1.5 meters above the ground, we should also calculate the additional time to reach this height after descending from maximum height. This total descent can be modeled using the equation:\[y = v_{0y}t + \frac{1}{2}gt^2\]Setting \(y = -1.5\) (to consider the downward path):\[-1.5 = 0 + \frac{1}{2}(9.8)t^2\]Solving this quadratic equation (using the positive root) provides the additional time to descend after reaching maximum height.

Step 3: Calculate Total Time of Flight

We already calculated the ascent time to the peak as 5.10 seconds (from Step 2). The descent portion involves solving the quadratic equation:\[0 = \frac{1}{2}(9.8)t^2 - 50t - 1.5\]Solving for \(t\) using the quadratic formula, we ignore the negative time value, and take the total time of ascent to peak and descent to ground:\[total\, t = 5.10 + Extra\_time\]

Step 4: Solve the Quadratic Equation

To find the quadratic equation solution, use:The quadratic formula is:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = \frac{1}{2}(9.8)\), \(b = 50\), \(c = -1.5\). Solving it entirely gives:\[total\, t \approx 10.2 \text{ seconds}\] (After evaluating both ascent and descent.)

Step 5: Calculate the Range

The range of the projectile is the total horizontal distance covered. Use the horizontal velocity\(v_{0x}\) and the total time of flight:\[Range = v_{0x} \times total\, t\]Substituting the values:\[Range = 50\sqrt{3} \times 10.2 \approx 883 \text{ meters}\]

Key Concepts

Initial Velocity

When dealing with projectile motion, understanding the initial velocity is crucial. In our example, the projectile is launched at 100 m/sec at a 30° angle above the horizontal.
Breaking this down, we can calculate the horizontal and vertical components using trigonometric functions:

• Horizontal component (\(v_{0x}\) ): This is calculated using the cosine of the angle. Hence, \(v_{0x} = 100 \times \cos(30°) = 50\sqrt{3} \text{ m/s}\).
• Vertical component (\(v_{0y}\) ): This is calculated with the sine of the angle. Therefore, \(v_{0y} = 100 \times \sin(30°) = 50 \text{ m/s}\).

These components are essential as they allow us to determine how the projectile moves over time.

Time of Flight

The time of flight is how long the projectile stays in the air from launch until it hits the ground. It's influenced by several factors, including the initial velocity, the angle of launch, and the height from which it is fired.

In our problem, starting 1.5 m above the ground means we need to account for both the ascent (to the peak) and the descent (back down past the initial height and to the ground). Initially, the time to the maximum height is calculated when the vertical velocity becomes zero. This calculation is done using the initial vertical velocity and the gravitational pull:

• \( v = v_{0y} - gt\)
• At maximum height, \(v = 0\). Solving gives \(t = \frac{50}{9.8} \approx 5.10 \text{ seconds}\).

To determine how long it takes to fall back down to ground level, we solve a quadratic equation derived from the motion, taking into account the extra 1.5 m the projectile falls.

Quadratic Equation

The solution to calculate further descent time involves solving a quadratic equation. This equation models the additional time the projectile takes to descend after reaching the maximum height. The general form comes from plugging into the kinematic formula for motion:

• \(y = v_{0y}t + \frac{1}{2}gt^2\)
• Setting \(y = -1.5\) to consider the downward path from the peak.

Solving this quadratic equation involved using the quadratic formula:

• \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Where \(a = \frac{1}{2}(9.8),\ b = 50,\ c = -1.5\).

After solving, it results in the additional time beyond reaching maximum height. Adding this time to the ascent time gives the total time of flight.

Trigonometric Functions

In projectile motion, trigonometric functions are employed to break down the initial velocity into horizontal and vertical components. Our exercise demonstrates this with a projectile launched at 30°. The trigonometric uses are:

• Cosine for the horizontal component: \(\cos(30°) = \frac{\sqrt{3}}{2}\).
• Sine for the vertical component: \(\sin(30°) = \frac{1}{2}\).

These angles trigger specific geometric relationships that determine how fast and far the projectile moves in each direction. This understanding lends itself to various applications beyond this exercise, emphasizing the practical role of trigonometry in physics.