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 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.

Accepted Answer

$(a, b)$ Let the ball be projected up with an initial velocity $u$ . It passes through point $P$ at $P$ 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, $s = v_{0} t + \frac{1}{2} a t^{2}$ , we get,

$h = u t_{1} - \frac{1}{2} g t_{1}^{2} (upward motion) . . . (i) \begin{array}{l} \end{array}$

$h = u t_{2} - \frac{1}{2} g t_{2}^{2} (downward motion) . . . (ii) \begin{array}{l} \end{array}$

Hence, from $(i)$ and $(ii)$ , $u t_{1} - \frac{1}{2} g t_{1}^{2} = u t_{2} - \frac{1}{2} g t_{2}^{2}$ ,

$u (t_{2} - t_{1}) = \frac{1}{2} g (t_{2} - t_{1}) (t_{2} + t_{1}) \Rightarrow u = \frac{1}{2} g (t_{2} + t_{1})$ .

Substituting the value of $u$ in $(i)$ , we get,

$h = \frac{1}{2} g t_{1} t_{2}$ .

$(c)$ Using the relation, $\vec{v} = \vec{u} + \vec{a} t$ , we get,

$v_{p} = (\frac{g (t_{2} + t_{1})}{2}) - g t_{1} = \frac{g}{2} (t_{2} - t_{1})$ .

$(d)$ $(i)$ The maximum height reached by the ball from $A$ is given as,

$H_{\max} = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - {((\frac{1}{2}) g (t_{2} + t_{1}))}^{2}}{2 g} = \frac{g}{8} {(t_{2} + t_{1})}^{2}$ .

$(ii)$ The height reached by ball from point $P$ ,

$h^{'} = H_{\max} - h = \frac{g}{8} {(t_{1} + t_{2})}^{2} - \frac{1}{2} g t_{1} t_{2} = \frac{g}{8} {(t_{2} - t_{1})}^{2}$ .

$(e) (i)$ $A$ 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,

$< v > = \frac{Δ y}{Δ t} = \frac{h}{t_{1}} = \frac{(\frac{1}{2}) g t_{1} t_{2}}{t_{1}} = \frac{1}{2} g t_{2}$ .

(ii) The motion from $A$ to $P$ for time $t_{2}$ ,

average speed $= \frac{total distance}{total time}$

$= \frac{h + 2 h^{'}}{t_{2}} = \frac{\frac{1}{2} g t_{1} t_{2} + 2 (\frac{g}{8} {(t_{2} - t_{1})}^{2})}{t_{2}} = \frac{g (t_{1}^{2} + t_{2}^{2})}{4 t_{2}}$ .

The motion from $A$ to $P$ for time $t$ ,

average velocity,

$< \vec{v} > = \frac{Δ \vec{y}}{Δ \vec{t}} = \frac{h}{t_{2}} = \frac{(\frac{1}{2}) g t_{1} t_{2}}{t_{2}} = \frac{1}{2} g t_{1}$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 28: ILLUSTRATION 4.26

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 28: ILLUSTRATION 4.26

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 28: ILLUSTRATION 4.26 with Hints & Solutions

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