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18 Jun 2024, 02:25
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Bunuel
Help me understand. If you something is thrown vertically upward and it has reached it max height then it will start falling down. If at t=3 its reaches the max height of 150 ft. then in the another 2 seconds it should start to fall down ?
if this is the case then height in 2 seconds is 134 ft. therefore the differnce between 150 and 134 will give me the actual height of the object.
can you please tell me where i am going wrong with this approach ? Bunuel KarishmaB
18 Jun 2024, 02:43
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adityakaregamba
I think you are misunderstanding the problem. The object reaches its maximum height from the ground 3 seconds after being thrown upward. The question asks for the height from the ground of the object 2 seconds after reaching this maximum, so 5 seconds after it was thrown. Substituting t = 5 into the equation should give this height.
18 Jun 2024, 07:06
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IEsailor
\(h = -16 (t -3)^2 + 150\)
We need to find the height of the object 2 seconds after it reaches its maximum height. When does it reach its maximum height?
Note that we need to maximize: \(- 16 (t -3)^2 + 150\) which is a quadratic.
It will look something like \(-16t^2 + 96t ...\)
Since co-efficient of t^2 is negative, it is downward opening parabola and maximum value of the expression will be at
\(t = \frac{-b}{2a} = \frac{-96}{2*(-16)} = 3\)
2 secs after reaching maximum height, the object is coming back and will be at the same point at which it is at t = 1.
\(h = -16 (1 -3)^2+ 150 = 86\)
Answer (B)
Check quadratics and parabola here: https://youtu.be/QOSVZ7JLuH0
