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3. A ball is thrown vertically downward from the edge of a high cliff with an initial velocity of 25ms ¹. (a) How fast is it moving after 1.5 s? (b) How far has it moved after 1.5 s? (take g=10m/s²)



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Final answer:

The ball will be moving at 40 m/s after 1.5 seconds, and it will have moved 48.75 meters by then when thrown downward with an initial velocity of 25 m/s from a high cliff, considering an acceleration due to gravity of 10 m/s².

Step-by-step explanation:

Understanding Free Fall and Velocity

A ball is thrown vertically downward from the edge of a high cliff with an initial velocity of 25 m/s. To determine how fast it is moving after 1.5 seconds, we use the equation of motion for objects in free fall:
v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (10 m/s²), and t is the time elapsed.

(a) To find the final velocity after 1.5 s:
v = 25 m/s + (10 m/s² * 1.5 s) = 25 m/s + 15 m/s = 40 m/s.

For the distance moved, we use the formula:
s = ut + 0.5gt², where s is the displacement.

(b) To find the displacement after 1.5 s:
s = (25 m/s * 1.5 s) + 0.5 * 10 m/s² * (1.5 s)² = 37.5 m + 0.5 * 10 * 2.25 = 37.5 m + 11.25 m = 48.75 m.

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