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if a satellite is orbiting earth at constant speed, select one: a. its orbit is unbound. b. its orbit is an elongated ellipse. c. its orbit is circular. d. it is just under its escape velocity.

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Answer:

The ball was thrown from a height that can be calculated using the kinematic equation:

h = vi * t + 0.5 * a * t^2

Where:

h = height from which the ball was thrown

vi = initial vertical velocity = 0 m/s (thrown horizontally)

a = acceleration due to gravity = -9.8 m/s^2

t = time taken to hit the ground

We can use the horizontal distance traveled, d = 36.31 m, to calculate the time:

d = vi * t

t = d / vi

t = 36.31 / 29.59

Now we can use t to find h:

h = vi * t + 0.5 * a * t^2

h = 0 * t + 0.5 * (-9.8) * t^2

h = 4.8 * t^2

h = 4.8 * (36.31 / 29.59)^2

h = 10.76 m

So, the ball was thrown from a height of 10.76 m.

Step-by-step explanation:

User Caleb Shay
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