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 = gravityInformation 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.
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