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16 April, 11:28

A 2.25 kg ball experiences a net force of 965 N up a ramp as shown. Once the ball reaches the top of the ramp, the force no longer acts. The force acts over a distance of 1.50 m on the ramp. Find the horizontal distance x that the ball travels before it hits the deck. The top of the ramp is 4.50 meters above the deck below.

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  1. 16 April, 12:32
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    consider the motion of ball on the ramp ::

    a = net acceleration on ramp = Net force / mass = 965/2.25 = 428.89 m/s2

    d = length of the ramp = 1.50 m

    Vi = initial velocity = 0

    Vf = final velocity at the top of the ramp

    Using the equation

    Vf2 = Vi2 + 2 a d

    Vf2 = 02 + 2 (428.89) (1.50)

    Vf = 35.87 m/s

    consider the motion of ball after it leaves the ramp ::

    Vix = Vf Cos28 = component of velocity in X-direction = 35.87 Cos28 = 31.67

    Viy = Vf Sin28 = component of velocity in Y-direction = 35.87 Sin28 = 16.84

    consider the vertical motion of the ball:

    Y = vertical displacement = height of top of ramp above the deck = - 4.50 m

    a = - 9.8

    Using the equation

    Y = Viy t + (0.5) a t2

    - 4.50 = 16.84 t + (0.5) (-9.8) t2

    t = 3.686

    distance travelled in X-direction is given as ::

    X = Vix t = 31.67 x 3.686 = 116.74 m
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