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Question: Anmol fired a Diwali rocket from the ground. A quadratic equation represents the height of a rocket ...

Anmol fired a Diwali rocket from the ground. A quadratic equation represents the height of a rocket from ground if the maximum height it reaches is 34 feet after 3 seconds it was fired. Which of the following represents the relation of h with t seconds after it was fired?
A) h(t)=−16(t−3)2+34h\left( t \right) = - 16{\left( {t - 3} \right)^2} + 34
B) h(t)=−16(t+3)2+34h\left( t \right) = - 16{\left( {t + 3} \right)^2} + 34
C) h(t)=16(t−3)2+34h\left( t \right) = 16{\left( {t - 3} \right)^2} + 34
D) h(t)=16(t+3)2+34h\left( t \right) = 16{\left( {t + 3} \right)^2} + 34

Explanation

Solution

It’s given as that a quadratic equation represents something. it means we have to form a quadratic equation with some variable then we’ll maximise it with respect to some variable. then use differentiation to solve this problem.

Complete step by step solution:
We need to check the options one by one.
Option A:
h(t)=−16(t−3)2+34h\left( t \right) = - 16{\left( {t - 3} \right)^2} + 34
Now differentiate it with respect to t.
h′(t)=−32(t−3)h'\left( t \right) = - 32\left( {t - 3} \right)
Now to know whether it satisfies the condition of maxima or minima differentiate it one more time and equate it to zero.
h′′(t)=−32<0h''\left( t \right) = - 32 < 0
h′(t)=0h'\left( t \right) = 0 at t=3.
Now let's check for maximum height at t=3.
h(t)=−16(t−3)2+34h\left( t \right) = - 16{\left( {t - 3} \right)^2} + 34

h\left( 3 \right) = 0 + 34 \\\ h\left( 3 \right) = 34 \\\ \end{gathered} $$ Thus it satisfies both the conditions. But I need to check for other options. Option B: $$h\left( t \right) = - 16{\left( {t + 3} \right)^2} + 34$$ Now differentiate it with respect to t, $$h'\left( t \right) = - 32\left( {t + 3} \right)$$ Now to know whether it satisfies the condition of maxima or minima differentiate it one more time and equate it to zero. $$h''\left( t \right) = - 32 < 0$$ $$h'\left( t \right) = 0$$ at t=-3, which is not possible. Time is never negative. So option B is cancelled. Same for option D also. Option C: $$h\left( t \right) = 16{\left( {t - 3} \right)^2} + 34$$ Now differentiate it with respect to t. $$h'\left( t \right) = 32\left( {t + 3} \right)$$ Now to know whether it satisfies the condition of maxima or minima differentiate it one more time and equate it to zero. $$h''\left( t \right) = 32 > 0$$ But this is not maximum height. Thus this option also eliminates. **Note:** Double differentiate will help you in finding the point of maxima or minima. This can also be done by single differentiation. Double differentiation tells us about the concavity or convexity of the curve. whereas single differentiation talks about the slope. And with slope, we can think of increasing-decreasing of a curve.