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Question: The function \(h(t) = - 16{t^2} + 80t\) represents the height of the baseball over time. How long do...

The function h(t)=−16t2+80th(t) = - 16{t^2} + 80t represents the height of the baseball over time. How long do you think the ball will be in the air?

Explanation

Solution

h(t)=−16t2+80t+0h(t) = - 16{t^2} + 80t + 0 seems a small amount off. This implies we are releasing the ball from a height of zero feet. Then you have to find the value of tt when height is again zero feet after releasing .

Complete step-by-step solution:
It is given that the function h(t)=−16t2+80th(t) = - 16{t^2} + 80t represents the height of the baseball over time.
We are solving for the tt when h(t)=0h(t) = 0, therefore we are going to set our function −16t2+80t=0 - 16{t^2} + 80t = 0
Since this is often a quadratic with the CC term missing, (ax2+bx+c)(a{x^2} + bx + c)
We will solve by factoring.
⇒−16t2+80t=0\Rightarrow - 16{t^2} + 80t = 0
For finding roots of the original equation, we have to use quadratic formula i.e.,
−b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Now identify a,b,ca,b,c from the original equation given below,
⇒−16t2+80t=0\Rightarrow - 16{t^2} + 80t = 0
⇒a=−16 ⇒b=80 ⇒c=0  \Rightarrow a = - 16 \\\ \Rightarrow b = 80 \\\ \Rightarrow c = 0 \\\
Put these values into the formula of finding the roots of quadratic equations,
⇒x=−80±802−80∗(−16)∗(0)2∗(−16)\Rightarrow x = \dfrac{{ - 80 \pm \sqrt {{{80}^2} - 80*( - 16)*(0)} }}{{2*( - 16)}}
After simplifying and by evaluating exponents and square root of the above equation we get the following simplified expression,
⇒x=−80±80−32\Rightarrow x = \dfrac{{ - 80 \pm 80}}{{ - 32}}
To find the roots of the equations , separate the particular equation into its corresponding parts : one part with the plus sign and the other with the minus sign ,
⇒x1=−80+80−32 and ⇒x2=−80−80−32  \Rightarrow {x_1} = \dfrac{{ - 80 + 80}}{{ - 32}} \\\ and \\\ \Rightarrow {x_2} = \dfrac{{ - 80 - 80}}{{ - 32}} \\\
Simplify and then isolate xxto find its corresponding solutions!
⇒x1=0 and ⇒x2=5  \Rightarrow {x_1} = 0 \\\ and \\\ \Rightarrow {x_2} = 5 \\\
So , the ball is on the bottom a time , t=0t = 0 (right before you throw it) and goes up and comes backpedal and hits the bottom 55 seconds later, t=5t = 5 .

Therefore , the ball will be in the air for 5 second.

Note: The ball is on the bottom a time , t=0t = 0 (right before you threw it) . This implies we are releasing the ball from a height of zero feet and goes up and comes backpedal and hits the bottom 55 seconds later, t=5t = 5 .