Answer to Question #103636 in Mechanics | Relativity for Joy Fapohunda

Question

A device launches ballons at a tangent with speed 12ms^-1. The tangent is 14m away at the same elevation the side of the fence. How can this be accomplished with gravity = 9.8ms^-2

Answer 1

As per the question,

The speed of the balloon is (u)=12m/sec

distance d= 14m

Acceleration due to gravity(g)=9.8 $m/sec^2$

let v is the final velocity, which will be zero at the top of the projection.

Now, $v^2=u^2-2gh$

$0=12^2-2\times9.8\times h$

$h=\dfrac{144}{19.6}=7.35m$

So, Maximum distance from the point side of the fence=$\sqrt{h^2+d^2}=\sqrt{7.35^2+14^2}$

=$\sqrt{250.02}$

$=15.8m$