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A ball is projected with initial velocity 50m/s at an angle of elevation of 37 degrees from the top of a cliff 55 m high. Calculate the total time the ball is in the air and the maximum horizontal distance covered.

Sagot :

The total time the ball is in the air is 1.47 s and the maximum horizontal distance covered is 58.7 m.

Time of motion of the ball

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

55 = (50sin37)t + ¹/₂(9.8)t²

55 = 30.1t + 4.9t²

4.9t² + 30.1t - 55 = 0

Solve using formula method;

t = 1.47 s

Horizontal distance of the ball

X = vxt

where;

  • vx is the horizontal velocity
  • t is time of motion

X = (50 x cos37) x 1.47

X = 58.7 m

Learn more about horizontal distance here: https://brainly.com/question/24784992

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