A cannonball is launched horizontally off a 65 m high castle wall with a speed of 40 m/s. how long will the cannonball be in the air before striking the ground?

Answers

Answer 1

The cannonball will be in the air a time of: 3.6421 s

The horizontal launch formula we will use and the procedure is:

t = √((2 * h) / g)

Where:

t = timeh = heightg = gravity

Information about the problem:

h= 65 mg = 9.8 m/s²v= 40 m/st =?

Applying the time formula we have:

t = √((2 * h) / g)

t = √((2 * 65 m) / 9.8 m/s²)

t = √(130 m / 9.8 m/s²)

t = √(13.2653 s²)

t = 3.6421 s

What is horizontal launch?

Is the motion described by an object that has been thrown horizontally with no angle of inclination at a certain height and considers the effect that the force of the earth's gravity has on the thrown object.

Learn more about horizontal launch at: brainly.com/question/24949996

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