A ball is thrown directly downward, with an initial speed of 6.10 m/s, from a height of 31.0 m. After what interval does the ball strike the ground?
To determine the time interval for the ball to strike the ground, we can use the kinematic equation for free fall:
\[ h = v_0t + \frac{1}{2}gt^2 \]
where:
- \( h \) is the height from which the ball is dropped (31.0 m),
- \( v_0 \) is the initial velocity (6.10 m/s),
- \( g \) is the acceleration due to gravity (-9.8 m/s²), and
- \( t \) is the time interval we want to find.
We can rearrange the equation to solve for \( t \):
\[ 0 = \frac{1}{2}gt^2 + v_0t - h \]
This is a quadratic equation in the form of \( at^2 + bt + c = 0 \), where:
- \( a = \frac{1}{2}g \),
- \( b = v_0 \), and
- \( c = -h \).
We can use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
After inserting the values for \( a \), \( b \), and \( c \), we can calculate \( t \). Taking the positive root will give us the time when the ball strikes the ground since time cannot be negative in this context.