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A ball is thrown straight down at 12.0 m/s from a 35.4 m -high building. how long does it take to reach the ground?

User Ekgren
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Final answer:

The ball takes 3.79 seconds to reach the ground.

Step-by-step explanation:

The time it takes for the ball to reach the ground can be found using the equation for vertical motion. We can assume that the initial vertical velocity is 12.0 m/s, the initial position is 35.4 m, and the final position is 0 m. By using the equation y = yo + voyt - 0.5 * g * t^2, where y is the final position, yo is the initial position, voy is the initial velocity, and g is the acceleration due to gravity, we can solve for t. Plugging in the given values, we get:

0 = 35.4 + 12.0t - 0.5 * 9.8 * t^2

This equation can be simplified to a quadratic equation -4.9t^2 + 12.0t + 35.4 = 0. You can solve this equation using the quadratic formula, and you will find two solutions: t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory (once on the way up and once on the way down), we take the longer solution, which is t = 3.79 s.

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