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Math Help - Initial velocity of projectile launced from above origin

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    Initial velocity of projectile launced from above origin

    I understand how to calculate the initial velocity of an object launched at an angle from the normal, but suppose i have a projectile launched at a 45 degree angle 10 meters above the X-axis and the projectile lands on the X-axis 100 meters from where it was launched. How do I calculate the initial velocity of the projectile in this hypothetical?
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  2. #2
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    Quote Originally Posted by Mr. Rogers View Post
    I understand how to calculate the initial velocity of an object launched at an angle from the normal, but suppose i have a projectile launched at a 45 degree angle 10 meters above the X-axis and the projectile lands on the X-axis 100 meters from where it was launched. How do I calculate the initial velocity of the projectile in this hypothetical?
    Motion in the horizontal direction:

    Initial velocity = \frac{u}{\sqrt{2}} m/s.
    Acceleration = 0 ~ \text{m/s}^2.
    Displacement = 100 m
    Therefore t = ..... (A)

    Motion in the vertical direction (take upwards as positive and take the origin to be where the object is launched):

    Initial velocity = \frac{u}{\sqrt{2}} m/s.
    Acceleration = 9.8 ~ \text{m/s}^2.
    Displacement = -10 m.
    t = expression found above (A).

    Substitute all this into x = ut + \frac{1}{2} a t^2 and solve for u.
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