Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (a)

MATH104 December 2016

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Question 02 (a)
A ball is thrown upward from a roof of a $50$ foot building with initial velocity of $96{\text{ft}}/{\text{sec}}$ . The height after $t$ seconds is given by $h(t)=50+96t-16t^{2}$ : Determine the maximum height that the ball reaches. Simplify your answer.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
At the beginning, the ball goes up and the direction of velocity is upward, velocity decrease gradually. After it decreases to $v=0$ , the ball starts to falling down, the direction of velocity becomes downward.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the hint, the highest point is reached when $v=0$ . To get the velocity, we find the derivative of height; $v(t)=h'(t)=(50+96t-16t^{2})'=96-32t$ . Set $v(t)=0$ , then we have $t=3$ By that time, $h(t)=h(3)=50+96\cdot 3-16\cdot 3^{2}=194$ Answer: $\color {blue}194({\text{ft}})$