A ball initially at rest is dropped from a window 18.6 m off the ground. How long does it take for the ball to reach the ground? Round your answer to 2 decimal places.
Solve this equation for t:
(1/2) g t^2 = H = 18.6 m
is this equation from kinematics?
You should have seen the equation in your text by now. It is the solution to Newton's second law of motion when the acceleration (g) is constant and the initial velocity is zero.
You can call it kinematics, mechanics or physics. When I attended college, kinematics meant something else.
To find the time it takes for the ball to reach the ground, we can use the formula for the time it takes an object to fall freely under gravity:
\(t = \sqrt{2h/g}\)
Where:
\(t\) is the time it takes to reach the ground,
\(h\) is the height of the fall,
and \(g\) is the acceleration due to gravity, which is approximately \(9.8 \, \text{m/s}^2\).
Now let's substitute the given values into the formula:
\(t = \sqrt{2 \cdot 18.6 \, \text{m} / 9.8 \, \text{m/s}^2}\)
Calculating this, we get:
\(t \approx \sqrt{3.795918367346939} \approx 1.95\) (rounded to two decimal places).
Therefore, it takes approximately 1.95 seconds for the ball to reach the ground.