A ball is thrown directly downward with an initial speed of 7.00 m/s, from a height of 30.4 m. After what time interval does it strike the ground?
solve for t when h=0
h = 30.4 - 7t - 4.9t^2
To find the time interval it takes for the ball to strike the ground, we can use the equation of motion for vertical motion:
\[ d = v_0t + \frac{1}{2}at^2 \]
where:
- \( d \) is the displacement (distance) the ball travels (in this case, the height of the ball from which it is thrown)
- \( v_0 \) is the initial velocity of the ball
- \( t \) is the time interval
- \( a \) is the acceleration due to gravity
In this case, the ball is thrown downward, so the initial velocity (\( v_0 \)) is negative (-7.00 m/s), since we're considering downward motion as negative. The displacement (\( d \)) is -30.4 m and the acceleration (\( a \)) due to gravity is approximately -9.81 m/s².
Plugging these values into the equation, we have:
\[ -30.4 = -7.00t + \frac{1}{2}(-9.81)t^2 \]
Simplifying the equation and rearranging it into a quadratic form, we get:
\[ -4.905t^2 - 7t - 30.4 = 0 \]
To solve for \( t \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, \( a = -4.905 \), \( b = -7 \), and \( c = -30.4 \). Plugging in the values:
\[ t = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(-4.905)(-30.4)}}{2(-4.905)} \]
Now we can calculate \( t \).