Ncert Exercises & Solutions

Q. 39 A cricket bowler releases the ball in two different ways

(a) giving it only horizontal velocity, and

(b) giving it horizontal velocity and a small downward velocity.

The horizontal speed, $v_{s}$ $v_{s}$ at the time of release, is the same. Both are released at a height of H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

Hint: During projectile motion, horizontal velocity remains unchanged.

Step 1: Find net velocity for part (a).

(a)

When the ball is given only horizontal velocity horizontal velæity at the time of release

(u_{x}) = v_{s}

$(u_{x}) = v_{s}$

During projectile motion, horizontal velocity remains unchanged,
Therefore,

v_{x} = u_{x} = v_{s}

$v_{x} = u_{x} = v_{s}$

In verticle direction,

\begin{matrix} v_{y}^{2} = u_{y}^{2} + 2 gH v_{y} = \sqrt{2 gH} \end{matrix}

$\begin{array}{l} v_{y}^{2} = u_{y}^{2} + 2 gH \\ v_{y} = \sqrt{2 gH} \end{array}$

$∴$ Resultant speed of the ball at the bottom.

$\begin{array}{l} v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ = \sqrt{v_{s}^{2} + 2 g H} \end{array}$

Step 2: Find the net velocity when a small downward velocity is given.

(b) When the ball is given horizontal velocity and a small downward velocity

Let the ball be given a small downward velocity u.
In horizontal direction

$v_{x}^{'} = u_{x} = v_{s}$
In vertical direction

$v_{v}^{' 2} = u^{2} + 2 g H$
or

$v_{y}^{'} = \sqrt{u^{2} + 2 gH}$

$∴$ Resultant speed of the ball at the bottom

$v^{'} = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{v_{s}^{2} + u^{2} + 2 gH} . . . (ii)$

Step 3: Compare the speeds.

From Eqs. (i) and (ii). we get v' > v

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