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A ball is thrown from an initial height of 1 meter with an initial upward velocity of 20 m/s. The ball's height h (in meters) after t seconds is given by the following

h = 1+20t-5t²
Find all values of t for which the ball's height is 11 meters.

User Matt Vukas
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1 Answer

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Final answer:

The values of t at which the ball's height is 11 meters are 0.54 s and 3.79 s, corresponding to times during the ball's ascent and descent respectively.

Step-by-step explanation:

To find all values of t for which the ball's height is 11 meters, we substitute h = 11 into the equation h = 1 + 20t - 5t² and solve for t. This results in a quadratic equation: 0 = -5t² + 20t + (1 - 11), which simplifies to 0 = -5t² + 20t - 10. By applying the quadratic formula, we find that the solutions to the equation are t = 0.54 s and t = 3.79 s. These represent the times when the ball is at a height of 11 meters on its way up and on its way down, respectively.

Using these times, we can understand the motion of the ball. It first reaches 11 meters at t = 0.54 s while ascending and then again at t = 3.79 s during its descent. Therefore, there are two instances in time when the ball is at the specific height of 11 meters.

User Tom Clelford
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