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2 years ago

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?

1 answer:

4 0

So, this is a problem where the accleration is not provided, since it is implied. The only acceleration is acceleration due to gravity (9.8 m/s)

The equation we will use for this problem is $V^2 =V_{0}^2 + 2a (X-X_0)$

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.

We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest. We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

$V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69$

So, the ball will go 24.69 meters up

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2 years ago

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