Physics

A projectile is fired from a height of h meters at a speed of 55m/s and at an angle of 10 degrees above the horizontal. The projectile hits a target at a horizontal distance of 80m from where it was fired and at a vertical height of 1.2m.

Find the time for which the projectile is in the air before it hits the target?

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  1. Horizontal problem:
    U = 55 cos 10 = 54.16 m/s
    It goes 80 meters horizontal at that speed
    so
    t = 80/54.16 = 1.48 seconds

    Are you sure that is all you want to know?
    I would ask you how high h was.

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  2. Just in case that should come up
    Vertical problem:

    final height = h + Vi t - 4.9 t^2

    1.2 = h + Vi t - 4.9 t^2

    we know Vi and t from the horizontal problem.

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  3. Oh, I never did Vi in fact
    Vi = 55 sin 10 = 9.55 m/s

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