Projectile Motion - Only given time and range

AI Thread Summary
To determine the launch angle and initial speed of a projectile given a range of 4.14 m and a time-of-flight of 0.854 s, one can utilize the independence of horizontal and vertical motion. The horizontal velocity can be calculated using the range and time, revealing that it remains constant throughout the flight. Similarly, the vertical motion can be analyzed using the time and gravitational acceleration to find the initial vertical speed. By combining both the horizontal and vertical components, one can derive the initial velocity vector's magnitude and direction. This approach effectively allows for the calculation of both the launch angle and initial speed without needing them explicitly provided.
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Homework Statement


If the range of the projectile is 4.14 m, the time-of-flight is T = 0.854 s, and air resistance is negligible, determine the following. (Assume that the acceleration due to gravity has a magnitude of g = 9.80 m/s2.)

(a) What is the launch angle of the projectile?

(b) What is the initial speed of the projectile?

Homework Equations


See below.

The Attempt at a Solution


I don't even know where to start. The book I have has equations that require either the launch angle or the initial speed, so I have no idea how I could do this without knowing one or the other. I couldn't use any of them since I wasn't given either the launch angle or the initial speed.

I guessed for a and b because I had no idea what to do.

This is the first physics class that I have ever taken, so I have no previous experience with physics, which makes it very confusing for me.
 
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Welcome to PF!

Although you have not been given either the initial speed or the launch angle, you have been given other equivalent information that will allow you to determine both. For example, you know that for a projectile, the motion in the horizontal (or "x") direction is totally independent of the motion in the vertical (or "y") direction. This independence holds true because the only force acting, namely gravity, is entirely in the vertical direction. Therefore, the only acceleration that can occur is in the vertical direction (recall from Newton's second law how the acceleration depends upon the force). Do you understand?

What this means is that there is no horizontal acceleration. Therefore, the horizontal velocity must be constant. It does not change -- whatever it was initially, is what it is throughout the flight. Furthermore you KNOW what it was initially. It may not be given to you explicitly, but you know implicitly that the horizontal speed must be whatever speed results in a distance traveled of 4.14 m in a time interval of 0.854 s. (This follows because the range of a projectile is defined as the horizontal distance it covers.) In other words, you can combine the range and the flight time in order to determine the horizontal or x-component of the velocity.

In a similar manner, you can combine the flight time with your knowledge of the vertical acceleration in order to figure out what initial vertical speed it must have had in order for it to stay in the air for 0.854 s before hitting the ground again. Once you've solved for that, you'll have both the vertical and the horizontal components of the velocity. Therefore, you'll know everything there is to know about the initial velocity vector, both its magnitude (which is being asked for in part (b)) and its direction (which is being asked for in part (a)).
 
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