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A ball is thrown straight up into the air at 49 m/s. How high does it go?

A. 122.5 m
B. 100.5 m
C. 111.5 m
D. 110.5 m

Answer :

Final answer:

The ball reaches a height of 122.5 meters.

Explanation:

To find the height reached by the ball, we can use the equation:

Final velocity squared = Initial velocity squared + 2 * acceleration * displacement

Since the ball is thrown straight up, its final velocity at the highest point is 0 m/s. The acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).

Substituting the given values into the equation:

0 = (49 m/s)^2 + 2 * (-9.8 m/s^2) * displacement

Simplifying the equation:

0 = 2401 m^2/s^2 - 19.6 m/s^2 * displacement

Now, let's solve for the displacement:

19.6 m/s^2 * displacement = 2401 m^2/s^2

displacement = 2401 m^2/s^2 / 19.6 m/s^2

displacement = 122.5 m

Therefore, the ball reaches a height of 122.5 meters.

Learn more about projectile motion here:

https://brainly.com/question/12860905

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