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29. A ball is launched from ground level with an initial upwards velocity of \( 20 \mathrm{~m} / \mathrm{s} \) and an initial horizontal velocity of \( 30 \mathrm{~m} / \mathrm{s} \). How far from its starting position does the ball land assuming the ground is level? \( \begin{array}{ll}\text { A) } 30 \mathrm{~m} & \text { E) } \\ \text { C) } 120 \mathrm{~m} & \text { D) } 60 \mathrm{~m}\end{array} \)

Ask by Powers Cross. in the United States
Feb 04,2025

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The ball lands approximately 120 meters from its starting position.

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To find out how far the ball lands from its starting position, we first need to determine the time it stays in the air. The vertical motion can be computed using the initial vertical velocity of \( 20 \, \text{m/s} \) and the acceleration due to gravity (\( g \approx 9.81 \, \text{m/s}^2 \)). The time to reach the highest point is given by \( t = \frac{v_{y}}{g} = \frac{20}{9.81} \approx 2.04\, \text{s} \). The total time in the air will be double that, so \( T \approx 4.08\, \text{s} \). Now, we can find the horizontal distance traveled using the formula: \[ \text{Distance} = \text{Horizontal Velocity} \times \text{Time} = 30 \, \text{m/s} \times 4.08 \, \text{s} \approx 122.4 \, \text{m} \] Rounding off gives us \( 120 \, \text{m} \), so the answer is \( \text{C) } 120 \text{ m} \). Fun fact: The combination of horizontal and vertical motions in projectile motion means that the ball travels along a parabolic path! This "jumping ball" is influenced equally by its initial velocities and the gravitational pull that tends to bring it back down. Who knew physics could be so entertaining?

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