**A boy, standing on the top of a building, throws a stone up with a velocity of 16 ms**

^{–1}in a direction making an angle of 30º with the horizontal. If the height of the building is 9.8 m, the velocity with which the stone will strike the ground will be approximately**(a) 17 ms**^{–1} at angle tan^{–1}(2/**√****3) with the horizontal **

**(b) 21 ms**^{–1} at angle tan^{–1}(–2/**√****3) with the horizontal**

**(c) 25 ms**^{–1} at angle tan^{–1}(–**√****3/2) with the horizontal **

**(d) 27 ms**^{–1}** ****at angle tan ^{–1}(–**

**√**

**3/2) with the horizontal**

**(e) 29 ms**^{–1} at angle tan^{–1}(–2/**√****3) with the horizontal **** **

The horizontal and vertical components of the velocity when the stone strikes the ground at C are to be found. The magnitude and direction of the velocity can then be obtained.

The horizontal component of velocity (u_{x}) is the same throughout the trajectory (since the gravitational pull is vertically downwards) and is given by

** u _{x} = v_{x} =**

**u cosθ**= 16 cos30° =

**8√3 ms**

^{–1}.The initial vertical component of velocity (at the point of projection A) is **u _{y}** =

**u sin**

**θ**= 16 sin30° = 8 ms

^{–1}. The vertical component (v

_{y}) of the final velocity (at the point C) is given by

v_{y} ^{2} = 8^{2} + 2×(–9.8)×(–9.8), making use of the standard relation *v ^{2} = u^{2} + 2as.*

[We have taken u which is vertically upwards as positive. The acceleration (due to gravity) and the displacement are downwards and hence should be negative]

Therefore, v_{y}^{2} = √(256.08) so that v_{y} = ± 16 ms^{–1}.

You have to take the *negative* value since v_{y} is *downwards*.

Therefore, **v _{y} = **

**– 16**

**m**

**s**

**.**

^{–1}The magnitude of the final velocity at the point C is

**v** = √(v_{x}^{2} + v_{y}^{2}) = √(8√3)^{2} + 16^{2}) = **21 m****s**** ^{–1}**, approximately.

The velocity ‘v’ is at an angle of tan^{–1}(v_{y}/v_{x}) = tan^{–1}(–16/8√3) with respect to the horizontal. ** **

So, the angle is **tan ^{–1}(–2/**

**√3)**and the correct option is (b).

Now, suppose in the above question, you were asked to find just the* speed* ( or the magnitude of the velocity) with which the stone strikes the ground at C. Your answer will be 21 ms^{–1} as you have already found out. But you can find it out a little more easily using the energy conservation method as follows:

The kinetic energy ½ mv^{2} of the stone at the moment of hitting the ground is equal to the sum of the initial kinetic energy ½ mu^{2 }and the initial gravitational potential energy mgh (with respect to the ground). Therefore we have

½ mv^{2 }= ½ m×16^{2} + m×9.8×9.8, from which v = 21 ms^{–1}

The following question will be quite simple if you remember the basic things in circular motion and elastic collisions ( and of course projectiles).

**A steel sphere A projected up with a velocity of 10 ms**^{–1} at an angle of 60**º with the horizontal, collides elastically with an identical steel sphere B located at the highest point of the trajectory of A. The sphere B is the bob of a simple pendulum. After the collision, if the sphere B just moves along a vertical circle, what is the length of the pendulum? ( Take g = 10 ms**^{–2})

**(a) 50 cm (b) 55 cm (c) 60 cm (d) 65 cm (e) 70 cm **

Many among you will be able to work this out easily. Once you have prepared thoroughly for an examination like the AP Physics Exam or a Medical and Engineering entrance Exam, you will be able to get the answer in less than a couple of minutes!

I will be back with the solution shortly, but try your best.

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