A ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where \(h\ =\ -3\left(t\ -\ 10\right)^2\ +\ 250\). What is the height of the ball once it reached its maximum height and then descended for 7 seconds
A) 96 feet
B) 103 feet
C) 164 feet
D) 223 feet
E) 250 feet
A ball thrown up in the air is at a height of h feet, t seconds after it was thrown,
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- Max@Math Revolution
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Solution: We know that \(h\ =\ -3\left(\ t\ -\ 10\right)^2\ +\ 250\)
We will first find the value for ‘t’ for which ‘h’ will be maximum.
For ‘h’ to be maximum, \(-3\left(\ t\ -\ 10\right)^2\ \) should be maximum. Since \(\left(\ t\ -\ 10\right)^2\ \) is a perfect square, therefore, \(\left(\ t\ -\ 10\right)^2\ \) ≥ 0.
But, \(\left(\ t\ -\ 10\right)^2\ \) will be ≤ 0 [By the property of reverse inequality]
So, for ‘h’ to be maximum \(-3\left(\ t\ -\ 10\right)^2\ \)= 0
=> \(-3\left(\ t\ -\ 10\right)^2\ \) = 0
=> \(\left(\ t\ -\ 10\right)^2\ \) = 0
=> (t − 10) = 0
=> t = 10.
‘7’ seconds after ball has reached maximum height ‘h’ at t = 10 + 7 = 17.
=> \(h\ =\ -3\left(\ t\ -\ 10\right)^2\ +\ 250\)
=> \(h\ =\ -3\left(\ 17\ -\ 10\right)^2\ +\ 250\)
=> h = −3 * 49 + 250
=> h = -147 + 250
=> h = 103 feet
Answer B
We will first find the value for ‘t’ for which ‘h’ will be maximum.
For ‘h’ to be maximum, \(-3\left(\ t\ -\ 10\right)^2\ \) should be maximum. Since \(\left(\ t\ -\ 10\right)^2\ \) is a perfect square, therefore, \(\left(\ t\ -\ 10\right)^2\ \) ≥ 0.
But, \(\left(\ t\ -\ 10\right)^2\ \) will be ≤ 0 [By the property of reverse inequality]
So, for ‘h’ to be maximum \(-3\left(\ t\ -\ 10\right)^2\ \)= 0
=> \(-3\left(\ t\ -\ 10\right)^2\ \) = 0
=> \(\left(\ t\ -\ 10\right)^2\ \) = 0
=> (t − 10) = 0
=> t = 10.
‘7’ seconds after ball has reached maximum height ‘h’ at t = 10 + 7 = 17.
=> \(h\ =\ -3\left(\ t\ -\ 10\right)^2\ +\ 250\)
=> \(h\ =\ -3\left(\ 17\ -\ 10\right)^2\ +\ 250\)
=> h = −3 * 49 + 250
=> h = -147 + 250
=> h = 103 feet
Answer B
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Solution:Max@Math Revolution wrote: ↑Mon Oct 19, 2020 10:32 amA ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where \(h\ =\ -3\left(t\ -\ 10\right)^2\ +\ 250\). What is the height of the ball once it reached its maximum height and then descended for 7 seconds
A) 96 feet
B) 103 feet
C) 164 feet
D) 223 feet
E) 250 feet
Since -3(t - 10)^2 will be either 0 or negative, we see that h is maximum when -3(t - 10)^2 is 0 and this term will be 0 when t = 10. In other words, the ball reached its maximum height at t = 10. After it descended for 7 seconds, t = 10 + 7 = 17 and the height at t = 17 is:
h = -3(17 - 10)^2 + 250 = -3(49) + 250 = -147 + 250 = 103 feet
Answer: B
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