An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-values at each 1 s from t = 0 s to t = 6 s determine vx- values at each 1s from t=0s to t=6s

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9⁰.

Horizontal component of velocity = 50 cos 36.9 = 39.98 m/s

The horizontal velocity in projectile motion is constant.

So value from t = 0 seconds to t = 6 seconds is a constant, which is equal to 39.98 m/s.

a) Displacement along X-axis

    Displacement, X =  *t

     X (t=0) = 39.98*0 = 0 m

     X (t=1) = 39.98*1 = 39.98 m

     X (t=2) = 39.98*2 = 79.96 m

     X (t=3) = 39.98*3 = 119.94 m

     X (t=4) = 39.98*4 = 159.92 m

     X (t=5) = 39.98*5 = 199.90 m

     X (t=6) = 39.98*6 = 239.88 m

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

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