physics

A toy plane p starts flying from point A along a straight horizontal line 20 m above ground level starting with zero initial velocity and acceleration 2 m /s² as shown. At the same instant, a man P throws a ball vertically upwards with initial velocitu 'u'. Ball touches(coming to rest) the base of the plane at point B of plane's journey when it is vertically above the man. 'S' is the distance of point B from point A. Just after the contact of ball with the plane, acceleration of plane increases to 4m/s². Find:
(1) initial velocity 'u' of ball.
(2) distance 's'
(3) distance between man and plane when the man catches the ball back.(g=10 m/s²)

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  1. well, first do the vertical problem, the ball

    how long does it take to fall 20 m?

    20 = (1/2) (10) t^2
    t^2 = 4
    t = 2 seconds

    so it will take to seconds to fall. By symmetry it will also take 2 seconds to rise and hit the plane

    how fast is it going when it hits the ground?
    u = a t
    = 10*2 = 20 m/s
    again by symmetry it will need 20 m/s initial velocity up to peak at 20 m

    Now work on the horizontal problem, the plane.

    it accelerates at 2 m/s^2 for 2 seconds
    s = (1/2) a t^2 = (1/2)(2) (4)
    s = 4 meters

    its speed then is
    v = a t = 2*2 = 4 m/s
    it goes on for another 2 seconds, accelerating faster while the ball falls
    d = Vi t + (1/2) a t^2
    d = 4(2) + (1/2)(4)(4)
    = 8 + 8
    = 16
    so
    the horizontal distance between man and plane is 16 meters when he catches the ball
    BUT
    the plane is still 20 meters up
    so
    total distance = sqrt(16^2+20^2)
    = 25.6 meters

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  2. Genius

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  3. I really like your logical explanation.
    Thank you so much.

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  4. Could have been more precise

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  5. You need to work upon it.

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