A ball is thrown directly downward, with an initial speed of 11m/s, from a height of 56m. After what time interval does the ball strike the ground?

To find the time interval it takes for the ball to strike the ground, we can use the equation of motion:

\[ h = v_i t + \frac{1}{2} g t^2 \]

where:
- \( h \) is the initial height (56m)
- \( v_i \) is the initial velocity (11m/s)
- \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \))
- \( t \) is the time interval we want to find

Since the ball is thrown directly downward, the initial velocity (\( v_i \)) is negative (-11m/s) because the negative direction is downward. Thus, we will substitute \( v_i = -11 \) in the equation.

Now, to find the time interval when the ball strikes the ground, we need to solve the equation for \( t \). Rearranging the equation, we get:

\[ \frac{1}{2} g t^2 - 11t + 56 = 0 \]

This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where \( a = \frac{1}{2} g \), \( b = -11 \), and \( c = 56 \). We can solve this equation using the quadratic formula:

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

Substituting the values, we have:

\[ t = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(\frac{1}{2} g)(56)}}{2(\frac{1}{2} g)} \]

Simplifying, we get:

\[ t = \frac{11 \pm \sqrt{121 - (4)(\frac{1}{2})(9.8)(56)}}{9.8} \]

Evaluating the expression under the square root, we have:

\[ \sqrt{121 - (4)(\frac{1}{2})(9.8)(56)} \approx \sqrt{121 - 1090.4} \approx \sqrt{-969.4} \]

Since the square root of a negative number is undefined in this context, it means that the ball will not strike the ground. However, we can still find the time when the ball reaches its maximum height by using the equation:

\[ t = \frac{v_f - v_i}{-g} \]

where \( v_f \) is the final velocity, which is 0 at the maximum height.

Substituting the values, we get:

\[ t = \frac{0 - (-11)}{-9.8} \]

Simplifying, we find:

\[ t \approx \frac{11}{9.8} \approx 1.12 \, \text{s} \]

Therefore, the ball takes approximately 1.12 seconds to reach its maximum height, but it does not strike the ground.

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