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    gamachan's Avatar
    gamachan Posts: 1, Reputation: 1
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    #1

    Nov 19, 2008, 06:36 PM
    Projectil motion
    Well, I have tried to resolve this problem... I can't do it... the problem is in English, I'm not a English speaker, but I understand something...

    A ball is thrown at angles 45º and 30º above the horizontal with the same initial speed. What multiple of the horizontal range at angle 30º is that at angle 45º?

    The answer is 2/√3... (this isn't a homework)

    I would like that you tell me how you did it...
    Sorry, my english is horrible
    Thank you very much..
    harum's Avatar
    harum Posts: 339, Reputation: 27
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    #2

    Nov 19, 2008, 07:15 PM

    Well, have you read anything about projectile motion? Do you have a physics textbook handy? This is a straightforward problem requiring knowledge of just a few definitions. When you throw a ball at an angle above horizon, its trajectory is parabola, with the gravity being the only force acting on the ball. It is always assumed in these problems that there is no air drag. Gravitational force is directed down and is perpendicular to the horizon. This means that horizontal component of the velocity, Vh, does not change, the vertical component, Vv, changes as a function of time: Vv(t) = Vov - gt, where V(t) is a vertical component, Vov - initial vertical component, g - free fall acceleration 9.8 m/s^2. Find how long it will take for the ball to travel to the highest point of its trajectory from starting height of 0 using that equation. Than remember that the same time is necessary to fall back to the surface from the highest point. This will give you total travel time, from which you can find total horizontal displacement for both angles (do it separately for each angle): S=Vh*T, where T is total travel time. Do not worry that you do not have the value for the initial speed. You have the angles, and this is enough. Remember that both horizontal and vertical components of the ball's velocity will be different for different angles. Also remember: Vh=V*cos(alpha), and Vv=V*sin(alpha). Do you understand why? HTH
    visharad's Avatar
    visharad Posts: 47, Reputation: 4
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    #3

    Nov 20, 2008, 06:04 AM

    The question says same initial speed.
    Let initial speed = u,
    launch angle = theta
    Do you know the formula for range?
    Do the following:-
    1. Find range for theta = 45 deg. You will get answer in terms of u.
    2. Find range for theta = 30 deg. You will find answer in terms of u.
    3. Divide answer of 1 by answer of 2.

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