Answer:
The time the catcher have to get in position to catch the ball before it hits the ground is nearly 5 seconds.
Step-by-step explanation:
Given:
The variation of height of a ball above ground with time 't' is given as:
[tex]P(T)=80T-16T^2+3.5[/tex]
In order to reach the ground, the height will be 0 feet.
So, plugging in 0 for 'P(T)' and solving for time 'T', we get:
[tex]P(T)=0\\80T-16T^2+3.5=0\\16T^2-80T-3.5=0[/tex]
Solving for 'T' using the quadratic formula:
For a quadratic equation [tex]ax^2+bx+c=0[/tex], the solutions are:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
For the above equation, [tex]a=16,b=-80,c=-3.5[/tex]
Therefore, the values of 'T' are:
[tex]T=\frac{-(80)\pm\sqrt{(-80)^2-4(16)(-3.5)}}{2(16)}\\\\\textrm{On solving, we get:}\\\\T=5.04\ s\ or\ T=-0.04\ s[/tex]
Negative time is neglected as time can't be negative.
Therefore, the time the catcher have to get in position to catch the ball before it hits the ground is nearly 5 seconds.
