A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?

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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Hello!

Because nothing is specified, I suppose that the vertical shot is made (formally) from the height 0. Also we'll ignore air resistance.

Then the height of the marble during the first shot is

`H_1(t) = V_0*t-(g*t^2)/2,`

where `V_0` is the initial velocity of a marble.

This height will be at a maximum at `t_1=V_0/g,` and the maximum height will be

`H_1(t_1)=(V_0^2)/(g)-(V_0^2)/(2g)=V_0^2/(2g).`

And by the conditions this is 9m, so `V_0=sqrt(18g).`

Now turn the gun horizontally. The initial velocity will remain the same (it is the property of the system gun+marble).

The height in this case will be

`H_2(t)=1.8-(g*t^2)/2.`

Marble hit the ground when `H_2(t_2)=0,` i.e. `t_2=sqrt(3.6/g).` The horizontal distance is `L(t)=V_0*t,` and its maximum is at `t=t_2.`

So the range in question is

`V_0*t_2=V_0*sqrt(3.6/g) = sqrt(18g)*sqrt(3.6/g) = sqrt(18*3.6) approx` 8.1 (m).

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