The bullet is shot forward with a horizontal velocity $u_x$ . It takes a time t to fall a vertical distance y and at the same time travels a horizontal distance x.

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance y using the equation,

$y=u_yt+\frac{1}{2} gt^2$

Substitute 0 m/s for $u_y$ , 9.81 m/s²for g and 1.5 m for y.

$y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s$

The horizontal distance traveled by the bullet is given by,

$x=u_xt$

Substitute 500 m/s for $u_x$ and 0.5530s for t.

$x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m$

The bullet travels a distance of 276.5 m.

5 0

3 years ago

A high school physics student is sitting in a seat reading this question. The magnitude of the force with which the seat is push

Dennis_Churaev [7]

The force pushing down is the force of Gravity. On a chair it is in perfect balance with the force pushing up (the normal force)

in terms of magnitude
FN = FG = mg

the forces are in opposite direction

hope this helps

4 0

3 years ago

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