Question

A shell is fired vertically upwards with a velocity V1 from the trolley moving horizontally with velocity V2. A person on the ground observes the motion of the shell as parabola. Its horizontal range is Ans; 2V1V2/g

Answer 1

Solution :-

The Horizontal Range equals = 2V1V2 /g

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Answer 2

A shell is fired vertically upward with a velocity $V_1$ from the trolley moving horizontally with velocity $V_2$.

we have to find out Range of parabolic motion of shell.

we know, Range of oblique projectile $R=\frac{2(usin\theta)(cos\theta)}{g}$

here,$usin\theta$=vertical component of velocity

$ucos\theta$=horizontal component of velocity.

in question, vertical component is $V_1$

horizontal component is $V_2$

hence, range =$\frac{2V_1V_2}{g}$

hence, A person on the ground observes the motion of the shell as parabola. Its horizontal range is $\frac{2V_1V_2}{g}$