SOLUTION: When a ball is thrown upward at a speed of 19 m/s, its height s above the ground (in meters) after t seconds is given by the formula s=19t-4.9t^2 Find the height of the ball af

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Question 483292: When a ball is thrown upward at a speed of 19 m/s, its height s above the ground (in meters) after t seconds is given by the formula
s=19t-4.9t^2
Find the height of the ball after 3 seconds
Found 2 solutions by bucky, mathstutor494:
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
You are given that the height of a ball that is thrown upward with an initial velocity of 19m/s can be calculated by using the equation:
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where t is the number of seconds after the ball is released.
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You are now asked to find the height of the ball after 3 seconds. You can do this simply by substituting 3 for t in the equation as follows:
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First multiply the 19 times 3 and you get 57. Substitute this into the equation and you have:
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Next square the 3 by recognizing that is and that is 9. So substitute 9 for and the equation then becomes:
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Multiply 4.9 times 9 and get 44.1
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Substitute this into the equation and you have:
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Finally, subtract the 44.1 from 57 and you have the answer:
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Don't forget that the answer is in the units of meters. So after throwing the ball with an initial velocity of 19 m/s, after 3 seconds the ball is 12.9 meters above the height at which is was released.
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As a little bit of an explanation of what this equation means, recognize that distance is rate (the 19 m/s) times time. So if gravity didn't work, you launch the ball upward and the distance it would be after t seconds is given by 19 times t.
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However, gravity does work on the ball. The effect of gravity is given by the second term in the equation. Since gravity is working downward, it has a negative or downward effect on the ball and that's why the second term in the equation has a negative or minus sign.
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If gravity didn't work, 3 seconds after launch, the ball would have been 57 meters above the launch point. But the effect of gravity has taken away 44.1 meters of the height ... that's the . And the result is that the ball is only 12.9 meters above the launch height.
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The math simply provides a way to help you understand and analyze the physics of the situation.
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Good luck and I hope this helps.
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Answer by mathstutor494(120) (Show Source):
You can put this solution on YOUR website!
Height of the ball after 3 seconds is achieved by substituting t=3 in below formula
s=19t-4.9t^2
s=19*3-4.9*3^2
s= 57 - 4.9*9
s= 57- 44.1
s= 12.9 m