The trajectory of a projectile launched at an angle of 45\deg (under the influence of gravity and air resistance) as a function of time, t, in seconds is described by the following pair of equations:
The correct answer and explanation is:
To determine the trajectory of a projectile launched at a 45° angle under the influence of gravity and air resistance, we typically express its position as a function of time using two parametric equations: x(t)=v02⋅t⋅e−ktx(t) = \frac{v_0}{\sqrt{2}} \cdot t \cdot e^{-kt} y(t)=v02⋅t⋅e−kt−12gt2y(t) = \frac{v_0}{\sqrt{2}} \cdot t \cdot e^{-kt} – \frac{1}{2}gt^2
Where:
- x(t)x(t) is the horizontal position at time tt
- y(t)y(t) is the vertical position at time tt
- v0v_0 is the initial velocity
- gg is the acceleration due to gravity (approximately 9.8 m/s²)
- kk is a constant representing air resistance
- tt is time in seconds
Correct Answer:
The correct form of the trajectory includes exponential decay terms due to air resistance. Without specific numerical values, we can’t pinpoint one final form, but the correct answer must reflect:
- A 45° launch implies equal horizontal and vertical components of initial velocity: v0x=v0y=v02v_{0x} = v_{0y} = \frac{v_0}{\sqrt{2}}
- Air resistance causes a decrease in both components over time, often modeled by an exponential decay factor like e−kte^{-kt}
So, a likely correct pair of equations is: x(t)=v02te−ktx(t) = \frac{v_0}{\sqrt{2}} t e^{-kt} y(t)=v02te−kt−12gt2y(t) = \frac{v_0}{\sqrt{2}} t e^{-kt} – \frac{1}{2}gt^2
Explanation (300 words):
Projectile motion under ideal conditions (no air resistance) is parabolic, with horizontal and vertical motions independent of each other. However, when air resistance is included, especially when it’s proportional to velocity (linear drag), the trajectory deviates significantly from a parabola and becomes more complex to model.
At a 45° launch angle, the initial velocity has equal horizontal and vertical components. In ideal conditions, both components remain constant in magnitude (horizontal) or decrease linearly with gravity (vertical). However, with air resistance, both components are affected. Air resistance acts in the opposite direction of motion, reducing the velocity magnitude exponentially over time. This is typically modeled using an exponential decay term such as e−kte^{-kt}, where kk is the air resistance coefficient.
The horizontal position x(t)x(t) decreases more slowly because only air resistance affects it, whereas the vertical position y(t)y(t) is influenced by both gravity and air resistance. The subtraction of 12gt2\frac{1}{2}gt^2 in the vertical component reflects the downward pull of gravity, which increases quadratically with time.
This model more accurately describes real-world trajectories like those of a baseball or bullet. The object rises and falls more steeply and travels a shorter horizontal distance than predicted by the no-drag model. Understanding this is critical in fields such as ballistics, sports science, and aerospace engineering, where air resistance can’t be ignored.
In conclusion, including air resistance transforms the simple parabolic path into a decaying, asymmetric trajectory, more reflective of reality.