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

Question #103636
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
1
Expert's answer
2020-02-24T11:00:01-0500

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"



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