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A ball is thrown horizontally from the top of a building 19 m high. The ball strikes the ground at a point 98.2 m from the base of the building. The acceleration due to gravity is 9.8 m/s².

Find the time the ball is in motion.

Note: The "x component of its velocity just before it strikes the ground" refers to the horizontal velocity, and the "y component of its velocity just before it strikes the ground" refers to the vertical velocity.

Answer :

Final answer:

The time the ball is in motion is approximately 2 seconds.

Explanation:

To find the time the ball is in motion, we can use the equation for vertical motion. The ball falls from a height of 19 m, so we can use the equation

h = ut + (1/2)gt^2,

where h is the height, u is the initial vertical velocity (0 m/s in this case), g is the acceleration due to gravity (9.8 m/s²), and t is the time. Substituting the values, we get

19 = (1/2)(9.8)t^2.

Solving for t, we find that the time the ball is in motion is approximately 2 seconds.

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