A marble is fired horizontally from a launching device attached to the edge of a tabletop which is 94.0 cm above the floor. The marble then strikes the floor 2.35 meters from the edge of the table.

What is the initial velocity of the marble as it leaves the launching device?

Time in the air of the marble using the equation y=1/2gt^2, re-arranged to make t the subject - t = √(2y/g). Therefore… t = √(0.94x2/9.81) = 0.4377… where g=9.81.
Using the equation x=Ut (the horizontal displacement) 2.35=u x 0.4377.. re-arrange to make u = x/t. Therefore.. u=2.35/0.4377 = 5.368m/s =5.37 = 3 significant figures

Answered by jackstattershal | 2024-06-10

The initial velocity of the marble as it leaves the launching device is 5.35 m/s ​ . ;

Answered by avantikar | 2024-06-12

The initial velocity of the marble as it leaves the launching device is approximately 5.36 m/s. This was determined by calculating the time it takes to fall from 94.0 cm and then using that time to find the horizontal speed. The marble travels 2.35 m horizontally during its fall.
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Answered by avantikar | 2024-09-30