Abaseball is thrown horizontally from a height of 9.60 m above the ground with a speed of 30.0 m/s. where is the ball after 1.40 s have clapsed?

The baseball will be  42 meters horizontally from where it was thrown and at the heigth of the ground.

Explanation:

this is a problem that involves basic mechanics in two directions, for the x axis we have a constant speed problem and for the y axis we have a constant aceleration problem, using the equations for each one of these we got:

where Xo and Yo are the inital position for x and y axis, in this case and , Vox and Voy are the initial speed for x and y axis, in this case and since the ball was throwns horizontally, and a is the acceleration in the y axis which is the gravitational aceleration, that means (the minus sign is because the gravitational aceleration is towards the negative y axis).

Replacing these values we have to equations that will help us find the position of the ball at any time we want:

and here we just have to replace the time value and solve, so:

and that's it, the ball is 42 meters horizontally from where it was throwns and at the heigth of the ground.

the device is called “ rocket engine”

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