Difficult Kinematic Range Problem

[0aNoMaLi7]

Kinematic Range Problem Help!

This problem is kicking my butt...

A ball is to be shot from level ground with a certain speed. The figure attached shows the range R it will have versus the launch angle 'theta'at which it can be launched. (The vertical axis is marked in increments of 20.0 m.) The choice of 'theta' determines the flight time. Let tmax represent the maximum flight time in seconds. What is the least speed the ball will have during its flight if 'theta' is chosen such that the flight time is 0.200tmax?

This problem is way too theoretical for me. I know that max range for a projectile is 45 degrees as dictated by the equation R= (V^2/g) x sin2'theta'. I could really use some direction. Thanks. This place rocks! I have told many of my classmates.

 

[Darien]

Call the initial velocity v_0

being launched at an angle, you have the following velocities in the x and y direction:
x: v_0 * cosA
y: v_0 * sinA

To determine the time it takes for the ball to rise and fall, you can use the instantaneous position and velocity formulas:
y(t) = 0 + v_0 * sinA * t - (1/2)gt^2
v_y = v_0 * sinA - gt

When the ball reaches its highest point, the y velocity should be 0, so:
v_0 sinA = gt

the total time the projectile spends is 2t, so:
t_max = 2* v_0/g * sinA

Noting that the velocity is just a vector composed of the x and y direction velocities, it shouldn't be too difficult from here

I'm new and too lazy to learn the latex math typesetting, but I hope this helps :)

 

[M98Ranger]

Dear student,

I understand that this kinematic range problem is challenging for you, but don't worry, I am here to help! Let's break down the problem and see if we can find a solution together.

First, let's review the given information. We know that the range of a projectile depends on its launch angle 'theta' and its initial speed. The maximum range is achieved at 45 degrees, as you correctly stated. We also know that the flight time is directly related to the launch angle, with a maximum flight time of tmax seconds.

Now, the question is asking for the least speed the ball will have during its flight if 'theta' is chosen such that the flight time is 0.200tmax. This means that we need to find the minimum speed that will still allow the ball to have a flight time of 0.200tmax seconds.

To solve this problem, we can use the equation for flight time: t = (2Vsin'theta')/g. We also know that the maximum flight time is tmax, so we can set up the following equation: tmax = (2Vsin'theta')/g.

Now, we can rearrange this equation to solve for V: V = (gtmax)/(2sin'theta'). This gives us the minimum speed that the ball must have in order to have a flight time of 0.200tmax seconds.

I hope this helps guide you in the right direction. Remember to always review the given information and use the appropriate equations to solve the problem. Keep up the good work and good luck! And thank you for recommending this resource to your classmates, we appreciate it.