Problem 37
Question
You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes \(4.0 \mathrm{~s}\) for the dart to land back at the barrel. What is the maximum horizontal range of your gun?
Step-by-Step Solution
Verified Answer
The maximum horizontal range is 55.44 m when fired at a 45-degree angle.
1Step 1: Calculate the Time of Flight
When the dart is shot straight up, it takes a total of 4 seconds for a round trip, meaning it takes 2 seconds to reach its maximum height. Therefore, the time of flight when the dart is shot in a projectile motion is also 4 seconds.
2Step 2: Determine Initial Vertical Velocity
Since the dart takes 2 seconds to reach its maximum height, we can use the equation of motion for vertical velocity: \( v = u + at \). At the maximum height, the final velocity \( v = 0 \), thus \( 0 = u + (-9.8 \times 2) \). Solving, we find \( u = 19.6 \text{ m/s} \).
3Step 3: Apply Motion Equations to Maximum Horizontal Range
The maximum horizontal range is given when the dart is launched at a 45-degree angle, so the initial velocity in both the vertical and horizontal directions is \( u_x = u_y = \frac{19.6}{\sqrt{2}} \). The range \( R \) can be calculated using \( R = u_x \times T \).
4Step 4: Calculate the Maximum Horizontal Range
Using the horizontal velocity \( u_x = \frac{19.6}{\sqrt{2}} = 13.86 \text{ m/s} \) and the total time of flight \( T = 4 \text{ s} \), the range is \( R = 13.86 \times 4 = 55.44 \text{ m} \).
Key Concepts
Horizontal RangeInitial VelocityTime of Flight
Horizontal Range
In projectile motion, the horizontal range refers to the total distance traveled by an object across a horizontal plane. When you shoot an object like a dart at an angle, its path follows a curve—a parabola, to be precise. This path usually has two significant components: vertical and horizontal. The horizontal range is the distance from the launch point to where the object lands.
For maximum horizontal range, you should aim to shoot the object at a 45-degree angle. This is because, at this angle, both vertical and horizontal components of initial velocity are equal, optimizing the range.
For maximum horizontal range, you should aim to shoot the object at a 45-degree angle. This is because, at this angle, both vertical and horizontal components of initial velocity are equal, optimizing the range.
- The formula to calculate the horizontal range is based on the initial horizontal velocity and the total time the object stays in motion.
- Mathematically, it's calculated as: \( R = u_x \times T \), where \( u_x \) is the initial horizontal velocity and \( T \) is the time of flight.
Initial Velocity
Initial velocity is the speed at which an object starts its journey in a specific direction. In projectile motion, this is really important because it influences both how far and how high the object will travel.
When considering maximum horizontal range, initial velocity plays a pivotal role. To achieve optimum range, you need to decompose the initial velocity into two components: horizontal and vertical.
When considering maximum horizontal range, initial velocity plays a pivotal role. To achieve optimum range, you need to decompose the initial velocity into two components: horizontal and vertical.
- Both components are derived from the overall initial velocity using trigonometric principles. If the angle is 45 degrees, then both components are equal.
- The initial vertical velocity in our dart example is calculated when the dart reaches its maximum height with a velocity of 0, using the equation \( v = u + at \).
Time of Flight
Time of flight is the duration that a projectile is in motion from the moment it is launched until it lands. This time directly affects how far the projectile will travel, as it stays in the air longer, it can cover more distance horizontally if started with a good initial velocity.
In our specific dart example, the time of flight was straightforward to calculate. Because the dart was shot straight up and took a total of 4 seconds to land, we know this is the same for any angle it could be shot at, provided the initial velocity remains constant.
In our specific dart example, the time of flight was straightforward to calculate. Because the dart was shot straight up and took a total of 4 seconds to land, we know this is the same for any angle it could be shot at, provided the initial velocity remains constant.
- The total time of flight (\( T \)) includes both the ascent and descent of the projectile.
- You often deal with symmetric projectile motion where ascent time equals descent time, making calculations easier.
Other exercises in this chapter
Problem 36
Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height if air resistance is neglib
View solution Problem 36
(II) You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun
View solution Problem 37
(II) A baseball is hit with a specd of 27.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ} .\) It lands on the flat roof of a \(13.0-\mathrm{m}\) -ta
View solution Problem 38
(II) A baseball is hit with a speed of 27.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ} .\) It lands on the flat roof of a \(13.0-\mathrm{m}\) -ta
View solution