Find the average velocity of a projectile between the instants it crosses half the maximum heght.It is projected with a speed u at an angle ø with horizontal.
Answers
y = u sin θ t - 1/2 g t² vy =
u sinθ - g t
x = u cos θ t
vx
= u cosθ
vy = 0 at
maximum height, at t =
T = u sinθ / g
maximum
height : H = u² Sin²θ /2g
The path
of the projectile is symmetric about the maximum height position. So the
time required to reach from H/2 to H and the time it takes to reach from H
again to H/2 are same.
time
to drop a height of H/2 from H with an initial vertical velocity of 0.
u² Sin²θ/4g = 1/2 g t^2
t = u sinθ / (g√2)
So we
need to find the average speed from t = t1 to t2
t1 = u sinθ/2g
* (2-√2) t2 = u sinθ/2g (2 + √2).
speed v
= √[vx² + vy²] = g * √[
(t - u sinθ/g)² + u² cosθ²/g² ]
let
w = (t - u
sinθ/g) / (u cosθ/g)
dw = g dt / (ucosθ)
Limits
of integration: w1 =
-1/√2 w2 = 1/√2
v = u
cosθ √(1+ w²)
[tex]Average\
speed\ v_{avg}= \frac{1}{t2-t1}\int\limits^{t2}_{t1} {v} \, dt\\\\
=\frac{1}{(u\ cos\theta\sqrt2/g)}\frac{u\
cos\theta}{g}\int\limits^{\frac{1}{\sqrt2}}_{-\frac{1}{\sqrt2}} {u\
cos\theta\sqrt{1+w^2}} \, dw\\\\ =\frac{u\ cos\theta}{\sqrt2}\ Tan^{-1}[ w
]^\frac{1}{\sqrt2}_\frac{-1}{\sqrt2}\\\\=\sqrt2\ u\ cos\theta\ Tan^{-1}[
\frac{1}{\sqrt2}][/tex]
====
If we are
looking for the average velocity vector v = vx i + vy j then, the
vertical component vy is positive before reaching the peak, and is negative
afterwards symmetrically. Hence the vertical component cancels out.
Average
velocity vector