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A ball is thrown downward from the top of a building that is h=20 mh = 20\text{ m} tall with an initial speed of v0=8 m/sv_0 = 8\text{ m/s} downward. Taking upward as positive and using g=10 m/s2g = 10\text{ m/s}^{2} for simplicity:

a) How long does it take for the ball to reach the ground?

b) What is the ball's velocity when it hits the ground?

Solution

Since the ball is thrown downward with initial speed 8 m/s8\text{ m/s}, and we're taking upward as positive, we have v0y=8 m/sv_{0y} = -8\text{ m/s}. Setting our origin at ground level, we have y0=20 my_0 = 20\text{ m} and the ball reaches the ground when y=0y = 0.

Let's visualize this problem. The ball starts at y0=20 my_0 = 20\text{ m} with an initial downward velocity of 8 m/s8\text{ m/s}:

Loading visualization…

a) To find the time to reach the ground, we use the position equation:

y=y0+v0yt12gt2y = y_0 + v_{0y}t - \frac{1}{2}gt^2

Substituting y=0y = 0, y0=20 my_0 = 20\text{ m}, v0y=8 m/sv_{0y} = -8\text{ m/s}, and g=10 m/s2g = 10\text{ m/s}^{2}:

0=20+(8)t12(10)t20 = 20 + (-8)t - \frac{1}{2}(10)t^2 0=208t5t20 = 20 - 8t - 5t^2 5t2+8t20=05t^2 + 8t - 20 = 0

Recall, the quadratic formula for solving these types of equations is given by:

t=b±b24ac2a. t = \dfrac{ -b \pm \sqrt{b^{2}-4ac}}{2a}.

Using the quadratic formula with a=5a = 5, b=8b = 8, and c=20c = -20:

t=8±64+40010=8±46410=8±21.5410t = \frac{-8 \pm \sqrt{64 + 400}}{10} = \frac{-8 \pm \sqrt{464}}{10} = \frac{-8 \pm 21.54}{10}

Taking the positive solution:

t=8+21.5410=13.5410=1....t = \frac{-8 + 21.54}{10} = \frac{13.54}{10} = 1. ...

... continued in the full lesson.

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