A ball is thrown directly downward, with an initial speed of 7.40 m/s, from a height of 31.0 m. After what time interval does the ball strike the ground?
I thought it would be 31 m / 7.40 = 4.18 seconds, but it's the wrong answer.
The acceleration of gravity, g = 9.81 m/s^2, has to have an effect on the answer.
Solve 0 = 31 - 7.40 t - (4.905)t^2
for t. Take the positive root of the quadratic equation.
t = [7.40 - sqrt (76.56 +588.6]/-9.81
= 1.87 s
To find the time interval it takes for the ball to strike the ground, we need to use the kinematic equation for vertical motion:
d = v₀t + (1/2)at²
where:
- d is the vertical distance traveled (in this case, 31.0 m)
- v₀ is the initial vertical velocity (7.40 m/s)
- t is the time interval we want to find
- a is the acceleration due to gravity (-9.8 m/s², considering downward motion)
We need to rearrange the equation to solve for t:
31.0 = 7.40t + (1/2)(-9.8)t²
Starting with the second term, we have:
(1/2)(-9.8)t² = -4.9t²
So, now the equation becomes:
31.0 = 7.40t - 4.9t²
Rearranging the equation, we get a quadratic equation:
4.9t² - 7.40t + 31.0 = 0
To solve this equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values for a, b, and c from our quadratic equation:
t = (-(-7.40) ± √((-7.40)² - 4 * 4.9 * 31.0)) / (2 * 4.9)
Simplifying further:
t = (7.40 ± √(54.76 - 607.6)) / 9.8
t = (7.40 ± √(-552.84)) / 9.8
Since we have a negative value inside the square root, it means that there are no real solutions to this equation. This means the ball does not hit the ground. It might be possible that there was a mistake in the initial calculation of the time. Please double-check the information given in the problem statement.