Question

(a) With what velocity $(v_{0})$ should a ball be projected horizontally from the top of a tower so that the horizontal distance on the ground is $η H$ , where $H$ is the height of the tower?

(b) Also determine the speed of the ball when it reaches the ground.

Accepted Answer

(a) Given, horizontal distance $= η H$ , where $H$ is the height of the tower.

We know that distance $=$ velocity $\times$ time.

Here the time involved will be the time of flight as we are considering projectile motion, which is given by $T = \sqrt{\frac{2 H}{g}}$ , where, $g$ is acceleration due to gravity.

So, $v_{0} T = η H$ .

$\Rightarrow v_{0} \sqrt{\frac{2 H}{g}} = η H \Rightarrow v_{0} = η \sqrt{\frac{g H}{2}}$

Hence, the velocity $v_{0}$ is $η \sqrt{\frac{g H}{2}}$ .

(b) During the projectile motion, the horizontal component of velocity remains same and its vertical component keeps on changing under the effect of gravity.

So horizontal speed, $v_{x} = v_{0}$ and vertical speed, $v_{y} = \sqrt{2 g H}$ (initial velocity is zero)
Total speed $v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

$= \sqrt{v_{0}^{2} + {(\sqrt{2 g H})}^{2}} = \sqrt{η^{2} \frac{g H}{2} + 2 g H} = \sqrt{(\frac{η^{2} + 4}{2}) g H}$

Hence, the speed of the ball when it reaches the ground is $\sqrt{(\frac{η^{2} + 4}{2}) g H}$ .

B M Sharma Solutions for Chapter: Kinematics II, Exercise 25: ILLUSTRATION 5.23

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 25: ILLUSTRATION 5.23

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 25: ILLUSTRATION 5.23 with Hints & Solutions

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