A ball at a height *H* is released from rest at the instant *t =* 0 and it strikes a fixed touch plate P, inclined at 45º with respect to the vertical, at the instant *t* = *t*_{1}. The point where the ball strikes the plate is at a height *h* (fig.). The collision between the ball and the plate is elastic. After the collision the ball continues its motion and strikes the ground at the instant *t* = *t*_{2}. Now, answer the following questions:

(a) If the height *h* where the ball strikes the plate is varied, how will you indicate, graphically, the variation of *t*_{1} with (*H* – *h*)? Use the coordinate axes as shown in the figure below to show the nature of variation.

(b) After the collision what is the nature of the path followed by the ball? Pick out your answer from the following three options by putting a tick (√) mark against the correct one:

Straight line path ……

Parabolic path ……

Arc of a circle …..

Justify your answer.

(c) Derive an expression for the total time of flight* t*_{2} of the ball.

(d) Determine the horizontal displacement of the ball when it strikes the ground.

** AP Physics C** aspirants should answer the following part also:

*H*determine the value of

*h*for which the time of flight

*t*

_{2}of the ball is

*maximum*.

*H*–

*h*suffered by the ball in time

*t*

_{1}is shown in the adjoining figure.

[The graph is obtained from the equation *H* – *h = *0 + ½ *gt*_{1}^{2 }and is parabolic]

(b) Just after the collision, the ball moves horizontally. Its horizontal velocity component remains unchanged and it picks up vertical velocity (downward) because of the gravitational pull. The path followed is therefore *parabolic* as in the case of any projectile having two dimensional motion.

(c) During the time *t*_{1} the ball suffers a vertical displacement (*H* – *h*) and hence we have

*H* – *h = *0 + (½) *g t*_{1}^{2}

This gives *t*_{1} = [2(*H* – *h*)*/g*]^{1/2}

Since the touch plate is inclined at 45º with the vertical and the collision is elastic, the ball moves horizontally just after the collision. Its horizontal velocity remains constant throughout its trajectory. But it has vertical acceleration *g *directed downwards and the time t’ taken for the vertical displacement *h* is given by

*h = *0 + (½) *g t*’^{2}

[Note that the initial vertical velocity is zero just after the hit at the touch plate].

This gives *t*’ = [2*h/g*]^{1/2}

The total time of flight* t*_{2} = *t*_{1} + *t*’ = [2(*H* – *h*)*/g*]^{1/2} + [2*h/g*]^{1/2}

Or, *t*_{2} = [2*/g*]^{1/2 }[(*H* – *h*)^{1/2} + *h*^{1/2}]

(d) The horizontal speed *v* of the ball after the collision at the touch plate is the same as the vertical speed just when the ball hits the touch plate. Therefore we have

*v = *[2g(*H* – *h*)]^{1/2}

[This follows from *v*^{2 }= 0^{ }+ 2g(*H* – *h*)]

The horizontal displacement *R* of the ball is given by

*R *= *vt*’ = [2g(*H* – *h*)]^{1/2}[2*h/g*]^{1/2} = 2[*h*(*H* – *h*)]^{1/2}

[You might have come across this equation earlier in fluid mechanics. This is the horizontal range of a liquid jet coming out through a side hole in a liquid tank (See the post dated 3^{rd} December 2007)].

(e) For a given height *H* the time of flight *t*_{2} will change when the height *h* of the touch plate is changed. the time of flight *t*_{2} will be *maximum* when *dt*_{2}/*dh = *0

Therefore, [2*/g*]^{1/2}[–(½)(*H* – *h*)^{–1/2 }+ (½)*h*^{–1/2}] = 0

This gives *H* – *h *= *h* from which *h* = *H*/2

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