Physics

There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum $=0$

It is the horizontal velocity which is uniform and hence $v_{av}=u_{x}=ucosθ$

For a general point:

Displacement in Y-direction:

$y=usinθ×t−2gt_{2} $

Displacement in X-direction:

$x=ucosθ×t$

Now in order to calculate average velocity:

Average Velocity = $TotaltimeNetDisplacement $

$R=(v_{o}cosθ_{o})(T_{f})=(v_{o}cosθ_{o})(2v_{o}sinθ_{o})/g$ ......(1)

Or, $R=gv_{o}_{2}sin2θ_{o} $ ......(2)

Equation 2 shows that for a given projectile velocity $v_{o}$, $R$ is maximum when $sin2θ_{o}$ is maximum, i.e. when $θ_{o}=45_{o}$.

The maximum horizontal range is, therefore

$R_{m}=gv_{0}_{2} $

Now, $v_{x}=v_{o}cosθ_{o}$

and $v_{y}=v_{o}sinθ_{o}−gt$

Since, at this point, $v_{y}=0$, we have:

$v_{o}sinθ_{o}−gt_{m}=0$

Or, $t_{m}=(v_{o}sinθ_{o})/g$

Therefore, time of flight $=T_{f}=2t_{m}−2(v_{o}sinθ_{o})/g$ because of symmetry of the parabolic path.

and $y=(v_{o}sinθ_{o})t−(1/2)gt_{2}$

The maximum height $h_{m}$ is given by:

$y=h_{m}=(v_{o}sinθ_{o})(gv_{o}sinθ_{o} )−2g (gv_{o}sinθ_{o} )_{2}$ (for $t=t_{m}$)

Or, $h_{m}=2g(v_{o}sinθ_{o})_{2} $

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