Respuesta :
Answer:
Part A: Initial height is 9 feet.
Part B: Maximum height is 13 feet.
Part C: Time taken is 0.5 s.
Step-by-step explanation:
Given:
The equation to represent the height of the ball is given as:
[tex]d(t)=-16t^2+16t+9[/tex]
Where, 't' is time passed in seconds.
Part A:
In order to find the initial height of the ball, the initial time must be 0 seconds.
So, plug in 0 for 't' and solve for d(0). This gives,
[tex]d(0)=-16\times 0+16\times 0+9\\d(0)=9\ ft[/tex]
Therefore, the height from which the ball was tossed is 9 feet.
Part B:
The maximum height occurs when the derivative of the given equation with time is 0 as at the relative maximum, the derivative of the function is always 0.
So, differentiating the above equation with respect to time, we get:
[tex]d'(t)=-16\frac{d}{dt}(t^2)+16\frac{d}{dt}(t)+\frac{d}{dt}9[/tex]
[tex]d'(t)=-16(2t)+16(1)+0[/tex]
[tex]d'(t)=-32t+16[/tex]
Now, at maximum height, the derivative is 0. So,
[tex]d'(t)=0\\\\-32t+16=0\\\\32t=16\\\\t=\frac{16}{32}=0.5\ s[/tex]
Therefore, the maximum height is obtained by plugging in 0.5 for 't' in the height equation. This gives,
[tex]d(0.5)=-16\times 0.5^2+16\times 0.5+9\\\\d(0.5)=-4+8+9=13\ ft[/tex]
Hence, the maximum height reached by the ball is 13 ft.
Part C:
The time for which the derivative was 0 is the time taken by the ball to reach maximum height.
Therefore, from part B, the time for maximum height is 0.5 s.
Answer:
Part A- 9ft
Part B- 13 ft
Part C - 1/2 second
Step-by-step explanation:
You MUST type in either 0.5 for part C or use the fraction option on the site and do 1/2
