# Difficult Kinematic Range Problem

1. Sep 12, 2004

### 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.

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Last edited by a moderator: Sep 12, 2004
2. Sep 12, 2004

### 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 :)