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Derivation of Velocity-Time Relation by Graphical Method

A ball is pro...

Question

A ball is projected vertically up such that it passes through a fixed point after a time $t_{1}$ and $t_{2}$ , respectively. Find
a.The height at which the point is located with respect to the point of projection
b. The speed of the projection of the ball.
c.The velocity of the ball at the time of passing through point P.
d. (i) The maximum height reached by the ball relative to the point of projection A and (ii) maximum height reached by the ball relative to point P under consideration.
e. The average speed and average velocity of the ball during the motion from A to P for the time $t_{1}$ and $t_{2}$ , respectively.

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Solution

a,b. Let the ball be projected up with an initial velocity u. It passes through point P at t = $t_{1}$ during its ascent and at t = $t^{2}$ during its descent.
For the motion of the ball from A to P,
s = +h, $v_{0}$ = u, a = -g, and t = $t_{1}$ and $t_{2}$
Substituting the above value, in s = $v_{0} t + (1 / 2) a t^{2}$ , we get
h = $u t_{1}$ - (1/2) $g t_{1}^{2}$ (upward motion)............(i)
= $u t_{2}$ - (1/2) $g t_{2}^{2}$ (downward motion)..........(ii)
Hence, from the above equations, $u t_{1}$ - (1/2) $g t_{1}^{2}$ = $u t_{2}$ - (1/2) $g t_{2}^{2}$
u( $t_{2} - t_{1}$ ) = $\frac{1}{2} g (t_{2} - t_{1}) (t_{2} + t_{1})$ $\Rightarrow$ $u = \frac{1}{2} g (t_{2} + t_{1})$
Substituting the value of u in (i) and (ii), we get h = (1/2)g $t_{1} t_{2}$
Solving the above equation,we have
h = $\frac{g t_{1} t_{2}}{2}$ and $v = \frac{g (t_{1} + t_{2})}{2}$
c. Using the relation $\to v = \to u + \to a t$
$v_{p} = (\frac{g (t_{2} + t_{1})}{2})$ - $g t_{1} = \frac{g}{2} (t_{2} - t_{1})$
d.(i). Maximum height reached by abll from the point of projection.
Using $v^{2} = u^{2} + 2 a s$
0 = ${(\frac{g}{2 (t_{2} + t_{1}})}^{2}$ - 2gH $\Rightarrow H_{m a x} = \frac{g}{8} (t_{1} + t_{2})^{2}$
(ii). Height reached by ball from point P,
$h^{'} = H_{m a x} - h = \frac{g}{8} (t_{1} + t_{2})^{2} - \frac{1}{2} g t_{1} t_{2}$ = $\frac{g}{8} (t_{2} - t_{1})^{2}$
e.(i). A to P for time $t_{1}$
The average speed and average velocity from A to P will be same as the particle is moving in straight line without changing the direction.
< v > = $\frac{Δ y}{Δ t} = \frac{h}{t_{1}} = \frac{(1 / 2) g t_{1} t_{2}}{t_{1}} = \frac{21}{2} t_{2}$
(ii) Motion from A to P for time $t_{2}$ .
Average speed = $\frac{T o t a l d i s t a n c e}{T o t a l t i m e}$
= $\frac{h + 2 h^{'}}{t_{2}}$
= $\frac{(1 / 2) g t_{1} t_{2} + 2 (\frac{g}{8} (t_{1} + t_{2})^{2})}{t_{2}}$
= $\frac{g (t_{1}^{2} + t_{2}^{2})}{4 t_{2}}$
Motion from A to P for time $t_{2}$
Average velocity < $\to v$ > = $\frac{Δ \to y}{Δ \to t} = \frac{h}{t_{2}} = \frac{(1 / 2) g t_{1} t_{2}}{t_{2}} = \frac{1}{2} g t_{1}$

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