A stone projected from ground with certain speed at a angle @ With horizontal attains maximum height H1,when it is projected with same speed at an angle @ with vertical attains height H2.The horizontal range of projectile is .?

Dear Student,

Max. height of a projectile: HMAX=v02sin2θ2g 
or,         H1=v02sin2@2g                     eq 1 

and  H2=v02sin2(90-@)2g  or  H2=v02cos2@2g                       eq. 2

By adding eq 1 & 2 we get

or , v02=2g(H1+H2)

Now, Total Time of Projectile motion

vertical displacement y=0      and vertical component of velocity= v0sin@ 
or, 0=v0sin@*T -1gT22
  T=2v0sin@g         eq. 3

Now horizontal range is given by using horizontal component of velocity i.e. v0cos@

or R=v0cos@*T

    from eq.3 we have

R=v0cos@ x 2 v0 sin@g or R = v02sin(2@)g                        as sin2@= 2sin@cos@   

This is same for case angle @ measured from virtical.

    putting in value of v02

R = 2g(H1+H2)sin(2@)g or R = 2(H1+H2)*sin(2@)


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