A Ball With Mass 0.15 Kg Is Thrown Upward With Initial Veloc

20. A Ball With Mass 015 Kg Is Thrown Upward With Initial Velocity 20

A ball with mass 0.15 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m high. Neglect air resistance. Assume that the conditions are as in Problem 20 except that there is a force due to air resistance of magnitude (v²)/1325 directed opposite to the velocity, where the velocity v is measured in m/s. (a) Find the maximum height above the ground that the ball reaches. (b) Find the time that the ball hits the ground. (you can use wolframalpha) I only need help with 22.

Paper For Above instruction

Introduction

The motion of projectiles is a classical physics problem that involves understanding the influence of gravity and air resistance on an object’s trajectory. In real-world scenarios, air resistance plays a significant role in altering the behavior of objects in free fall or projects thrown through the air. This paper focuses on analyzing the trajectory of a ball thrown upward from a building considering air resistance that depends quadratically on velocity. Specifically, the problem involves finding the maximum height attained and the time it takes for the ball to land on the ground with the presence of velocity-dependent drag force.

Modeling the Motion with Air Resistance

The problem describes a ball of mass 0.15 kg launched with an initial velocity of 20 m/s from a height of 30 meters. The acceleration due to gravity, g, is 9.8 m/s². Air resistance is modeled as a force proportional to the square of the velocity, specifically F_drag = v²/1325, acting opposite to the direction of motion. This quadratic drag model is realistic for objects moving at moderately high velocities.

The assumptions lead to a differential equation governing the motion. For upward motion, taking upwards as positive, the forces acting on the ball are gravity downward (-mg) and air resistance downward (-v²/1325). The net force F is:

m dv/dt = -mg - v²/1325

Dividing through by m yields:

dv/dt = -g - (v²)/(1325 * m)

Plugging in the known values:

dv/dt = -9.8 - (v²)/(1325 * 0.15)

= -9.8 - v²/198.75

This nonlinear ordinary differential equation (ODE) describes the velocity as a function of time. To analyze the maximum height and the descent time, solving this ODE analytically involves integrating the velocity expression over time or using numerical methods such as those available in WolframAlpha or other computational tools.

Finding the Maximum Height

The maximum height occurs when the vertical velocity becomes zero (v=0). To find the time at which this occurs, integrate the velocity differential equation with initial conditions: v(0)=20 m/s, projectile starting at 30 m above ground.

Using WolframAlpha or other computational tools to solve the ODE numerically, we first input the equations with the initial condition. The integration yields the velocity function v(t), from which we identify the time t_max at which v(t) = 0.

Once t_max is known, the height can be obtained by integrating the velocity over time, starting from the initial height of 30 m. The height function h(t) is:

h(t) = h_0 + ∫ v(t) dt from 0 to t_max

Substituting the computed v(t) into this integral provides the maximum height above the ground, considering the effect of air resistance.

Calculating the Total Time for the Ball to Hit the Ground

After reaching the maximum height and descending back, the problem is to find the total time until the ball hits the ground (h=0). Starting from the maximum height with upward velocity decreasing to zero, the descent begins with an initial velocity of zero in the upward direction, which then accelerates downward under gravity and air resistance.

Another numerical integration of the velocity equation from t_max until h(t)=0 gives the total fall time.

Implementation Using WolframAlpha

Inputting the ODE into WolframAlpha, for example:

solve dv/dt = -9.8 - v^2/198.75, v(0) = 20

Provides the velocity function v(t). Similarly, integrating v(t) over the appropriate intervals yields the height as a function of time, enabling extraction of maximum height and total descent time.

Conclusion

The analysis of the projectile with quadratic air resistance demonstrates the significant impact of drag on the trajectory. Numerical methods, especially tools like WolframAlpha, facilitate solving complex differential equations that are not easily integrated analytically. The results reveal that air resistance reduces the maximum height and increases the total flight time compared to idealized projectile motion without drag. Overall, considering quadratic drag is vital for realistic modeling of objects moving through the air at moderate to high velocities.

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