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A ball is thrown horizontally with an initial velocity of 40.0 m/s from the edge of a building of a certain height. The ball lands at a horizontal distance of 70.4m from the base of the building. How long was the ball in the air?

A. 17.1s
B. 0.54s
C.1.85s
D. 12.4s

1 Answer

1 vote

Final Answer:

The ball was in the air for approximately 1.76 seconds. Option A is answer.

Step-by-step explanation:

Identify the relevant equations:

The horizontal motion of the ball is described by the following equation:

x = v_x * t

where:

x is the horizontal distance traveled (70.4 m)

v_x is the initial horizontal velocity (40.0 m/s)

t is the time of flight

Solve for the time of flight:

Rearrange the equation to solve for t:

t = x / v_x

Substitute the known values:

t = 70.4 m / 40.0 m/s

Calculate the time of flight:

t ≈ 1.76 seconds

Therefore, the ball was in the air for approximately 1.76 seconds.

Option A is answer.

Option B (0.54s) is incorrect: This time is too short for the ball to travel the horizontal distance of 70.4 meters.

Option C (1.85s) is incorrect: This time is slightly longer than the correct answer but is still a reasonable estimate.

Option D (12.4s) is incorrect: This time is much too long for the ball to travel the horizontal distance of 70.4 meters.

User Vman
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