Question

I got the first part, which is 11.0m for height of ball thrown. But the second question for how long it takes for the ball to return to ground level, I'm stuck. I got 1.5 seconds and that was wrong. SOS

Answer 1

The ball reaches a height of approximately 11.0 meters, and it takes approximately 3.0 seconds for it to return to the ground level.

To find the height (h) that the ball reaches, you can use the following kinematic equation:

h = v₀² / (2g)

where:

- v₀ is the initial velocity (14.7 m/s),

- g is the acceleration due to gravity (9.8 m/s²).

Substitute these values into the equation:

h = (14.7 m/s)² / (2 * 9.8 m/s²)

Calculate the result:

h ≈ 216.09 m/s² / 19.6 m/s² ≈ 11.0 m

So, the height of the ball thrown is indeed 11.0 meters.

Next, to find the time (t) it takes for the ball to return to the ground, you can use the following kinematic equation:

h = v₀t - (1/2)gt²

Rearrange this equation to solve for t:

t² - (v₀/g)t - (2h/g) = 0

Substitute the known values:

t² - (14.7/9.8)t - (2 * 11.0/9.8) = 0

Now, solve this quadratic equation using the quadratic formula:

`\[ t = \frac{{-b \pm \sqrt{{b² - 4ac}}}}{{2a}} \]`

where a = 1, b = -14.7/9.8, and c = -2 * 11.0/9.8.

`\[ t = \frac{{1.5 \pm \sqrt{{11.23}}}}{2} \]`

`\[ t_1 = \frac{{1.5 + \sqrt{{11.23}}}}{2} \]`

`\[ t_2 = \frac{{1.5 - \sqrt{{11.23}}}}{2} \]`

Calculate the values and choose the positive root since time cannot be negative.

t ≈ 3.0 s

So, it will take approximately 3.0 seconds for the ball to return to the ground level.