A ball is thrown upward with an initial velocity of 35 meters per second from a cliff that is 80 meters high. The height of the ball is given by the quadratic equation h = -49t^2 + 35t + 140 where h is in meters and t is the time in seconds since the ball was thrown. Find the time that the ball will be 60 meters from the ground. Round your answer to the nearest tenth of a second.
PS: I know the answer is 9.0 seconds because it is a test question I got wrong, I need steps if anyone can help:)
To find the time when the ball is 60 meters from the ground, we need to solve the quadratic equation h = -49t^2 + 35t + 140 for t when h = 60.
1. Start with the given equation: h = -49t^2 + 35t + 140
2. Replace h with 60: 60 = -49t^2 + 35t + 140
3. Rearrange the equation to bring all terms to one side, so it becomes a quadratic equation: -49t^2 + 35t + 140 - 60 = 0
4. Combine like terms: -49t^2 + 35t + 80 = 0
You now have a quadratic equation in the form of at^2 + bt + c = 0, where a = -49, b = 35, and c = 80. To solve this equation, you can use the quadratic formula.
The quadratic formula is t = (-b ± √(b^2 - 4ac)) / 2a
5. Plug in the values for a, b, and c into the quadratic formula: t = (-(35) ± √((35)^2 - 4(-49)(80))) / (2(-49))
Simplify: t = (-35 ± √(1225 + 15680)) / (-98)
Simplify further: t = (-35 ± √(16905)) / (-98)
6. Calculate the square root of 16905: √(16905) ≈ 130.02
7. Substitute this value into the equation: t = (-35 ± 130.02) / (-98)
Now, let's calculate the two possible solutions for t.
8. Solution 1:
t = (-35 + 130.02) / (-98)
t ≈ 95.02 / -98
t ≈ -0.969
9. Solution 2:
t = (-35 - 130.02) / (-98)
t ≈ -165.02 / -98
t ≈ 1.68
Considering the context of the problem, we can discard the negative solution. The time when the ball will be 60 meters from the ground is approximately 1.68 seconds.
Therefore, the correct answer is not 9.0 seconds. It seems there may have been an error in the test question.
To find the time that the ball will be 60 meters from the ground, we need to solve the given quadratic equation for the height of the ball (h) when it is 60 meters.
The quadratic equation is h = -49t^2 + 35t + 140, where h is the height in meters and t is the time in seconds.
1. Substitute h = 60 into the equation:
60 = -49t^2 + 35t + 140
2. Rearrange the equation by moving all terms to one side:
0 = -49t^2 + 35t + 140 - 60
0 = -49t^2 + 35t + 80
3. We can now solve this quadratic equation for t. There are different methods to solve quadratic equations, but for this example, we will use factoring.
4. The equation is in the form ax^2 + bx + c = 0, where a = -49, b = 35, and c = 80.
5. Seek two numbers whose product is equal to ac (or -49 * 80 = -3920), and whose sum is equal to b (which is 35).
6. After trying different pairs, we find that the numbers -80 and 49 satisfy these conditions.
7. Rewrite the equation replacing the middle term (35t) with -80t + 49t:
0 = -49t^2 - 80t + 49t + 80
8. Group the terms:
0 = (-49t^2 - 80t) + (49t + 80)
9. Factor out common terms from each group:
0 = t(-49t - 80) + 1(49t + 80)
10. Factor out the common factors:
0 = t(-49t - 80) + 1(49t + 80)
11. Distribute the -1 to each term inside the parentheses:
0 = (-49t - 80)t + (49t + 80)
12. Simplify:
0 = -49t^2 - 80t + 49t + 80
13. Combine like terms:
0 = -49t^2 - 31t + 80
14. Now that the quadratic equation is factored, set each group equal to zero:
-49t^2 - 31t + 80 = 0
15. Solve for t using factoring, quadratic formula, or graphing calculator. In this case, factoring will be used:
-49t^2 - 31t + 80 = 0
(-7t + 16)(7t + 5) = 0
16. Set each factor equal to zero:
-7t + 16 = 0 or 7t + 5 = 0
17. Solve for t in each equation:
-7t = -16 or 7t = -5
t = -16/-7 or t = -5/7
t = 2.286 or t = -0.714
18. Since time cannot be negative in this context, we discard the negative value and keep the positive one.
Therefore, the time when the ball will be 60 meters from the ground is approximately 2.3 seconds (rounded to the nearest tenth of a second).