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?

This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

Expert Answers

An illustration of the letter 'A' in a speech bubbles


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


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


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).

Approved by eNotes Editorial Team