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?
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.
(The entire section contains 163 words.)
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