Question

The function h(t) = −16t2 + 32t + 24 models the height h, in feet, of a ball t seconds after it is thrown straight up into the air. What is the initial velocity and the initial height of the ball?

Answer 1

Answer:

See explanation for details

Step-by-step explanation:

To find the initial height of the ball, you need to plug in 0 in place of time (because this is the initial height). The first and second term become 0, leaving you with a height of 24. To find the speed of the ball, the first step is to figure out how high up it gets. If you set graph this equation or solve by completing the square, you will find that the peak of this ball's flight is 40 feet up, 1 second in. Since it starts at a height of 24, the ball goes up a total of 16 feet. The formula for calculating these types of things is $d=v_o t+ \frac{1}{2}at^2$, where $v_o$ represents the intial velocity, a represents acceleration, t represents time, and d represents distance. In this case, We can plug in our known numbers to get:

$16=v_o \cdot 1 + \frac{1}{2}\cdot 9.8\cdot 1$

$v_o=\frac{16}{4.9}\approx 3.265 m/s^2$. Hope this helps! Comment below for more details, or if this was unclear.

Answer 2

Answer:

C 32ft/s;24 ft

Step-by-step explanation:

I did it on the quiz