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

the answer would be c

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