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A ball is thrown directly downward with an initial speed of 6.43 m/s from a height of 38.1 m. After what interval does the ball strike the ground? The acceleration of gravity is 9.8 m/s².

a) 2.09 s
b) 1.47 s
c) 3.25 s
d) 0.84 s

1 Answer

1 vote

Final answer:

To find the time interval it takes for the ball to strike the ground, we can use the kinematic equation for vertical motion. Plugging in the given values and solving the quadratic equation, the time interval is approximately 0.90 seconds.

Step-by-step explanation:

To find the time interval it takes for the ball to strike the ground, we can use the kinematic equation for vertical motion:

d = v0t + 1/2gt2

where d is the initial height (38.1 m), v0 is the initial velocity (-6.43 m/s, negative because it is directed downward), g is the acceleration due to gravity (-9.8 m/s²), and t is the time interval we want to find.

Plugging in the given values, we get:

38.1 = -6.43t + 1/2(-9.8)(t2)

By rearranging the equation, we obtain a quadratic equation:

4.9t2 - 6.43t + 38.1 = 0

Using the quadratic formula: t = (-b ± sqrt(b2 - 4ac)) / 2a

where a = 4.9, b = -6.43, and c = 38.1, we can solve for t.

The two solutions for t are approximately 4.42 seconds and 0.90 seconds. Since we are looking for the time interval it takes for the ball to strike the ground, the correct answer is 0.90 seconds (Choice d).

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