**Solved Examples on Kinematics**

**Question 1:-**A certain computer hard disk drive is rated to withstand an acceleration of 100g without damage. Assuming the drive decelerates through a distance of 2 mm when it hits the ground, from how high can you drop the drive without running it?

**Concept:**

The height say *y* from which the disk falls freely, can be related to its final speed before collision say *v _{y}* and the initial speed

*v*

_{0y}as:

(*v _{y}*)

^{2}– (

*v*

_{0y})

^{2}= 2

*gy*

From this expression, we calculate the speed of the disk just before it hits the ground but in terms of height *y*.

If you assume that the initial speed of the disk before it hits the ground is *v*^{'}_{0y}, the final speed to be *v*'* _{y}* and the distance travelled during the collision is given by

*y*

^{'}, then the acceleration of the disk during collision is given using the relation: (

*v*'

*)*

_{y}^{2}– (

*v*

^{'}

_{0y})

^{2}= 2

*ay*

^{'}

Since the speed of the disk before collision is known from above, we can equate the limiting value of acceleration to account for the height *y* from which the disk can be thrown without damaging it.

**Solution:**

From the equation of kinematics, we have

*y* = *v*_{0y}*t* + ½ *at*^{2} (for *y*_{0} = 0)

*v _{y}*=

*v*

_{0y}+

*at*

Where *v _{y}* represents the final speed of the disk,

*a*is its acceleration,

*v*

_{0y}is the initial speed and is the time taken by the disk to attain the final speed

*v*.

_{y}From the second equation, the time *t* is given as: *t* = (*v _{y}*-

*v*

_{0y})

*/*

_{ }*a*

Substitute the value in first equation

*y* = *v*_{0y} [(*v*_{y}-*v*_{0y})/*a*] + ½ *a* [(*v*_{y}-*v*_{0y})/*a*]^{2}

= [(*v*_{y}-*v*_{0y})/*a*] [*v*_{0y} + ½ *a*[(*v*_{y}-*v*_{0y})/*a*]]

= [(*v*_{y}-*v*_{0y})/*a*] [(2* v*_{0y}+* v*_{y}-* v*_{0y})/2]

= [(*v*_{y}-*v*_{0y})/*a*] [(*v*_{y}+*v*_{0y})/2]

So,

2*ay* = (*v _{y}*-

*v*

_{0y})(

*v*+

_{y}*v*

_{0y})

= (*v _{y}*)

^{2}– (

*v*

_{0y})

^{2}

Since the disk accelerates under the action of gravity and we take the convention of downward motion to be positive, we can equate *a* = *g* in the equation above as:

(*v _{y}*)

^{2}– (

*v*

_{0y})

^{2}= 2

*gy*…… (1)

Substitute the given value of *v*_{0y} to be zero, the above equation can be written as:

(*v _{y}*)

^{2}– (0 m/s)

^{2}= 2

*gy*

(*v _{y}*)

^{2}= 2

*gy*…… (2)

Assume that the distance travelled by the disk during collision is *y*^{'}, the speed of the disk just before the contact is* v*^{'}_{0y}, and the final speed after the collision is *v*'* _{y}*, such that the disk experience an acceleration

*a*, then equation (1) can be written as:

(*v*'* _{y}*)

^{2}– (

*v*

^{'}

_{0y})

^{2}= 2

*ay*

^{'}

Since the disk will come to rest after colliding, hits final speed *v*^{'}* _{y}* is zero. Also the initial speed before collision is equal to

*v*

_{y}calculated above.

Given that the disk crushes to a depth of 2 mm, the height *y*^{'} can be equated to it.

Substitute the appropriate values in the equation above,

(0 m/s)^{2} – (*v*_{y})^{2} = 2*a *(2 mm)

Substitute the value of (*v*_{y})^{2} from equation (2),

-2*gy* = 2*a *(2mm)

Given that the limit to the deceleration that the disk can withstand during the collision is 100 g, we can equate *a* = 100 *g* as,

-2*gy* = 2(-100 g)(2mm)

*y* = 100 (2 mm) (10^{-3} m/1 mm)

So, *y* = 0.2 m

Therefore if the disk is thrown from height 0.2 m the disk will start to damage, therefore the person can throw the disk from any height which is less than 0.2 m and avoid damaging it.

**Question 2:- **A high-performance jet plane, practicing radar avoidance maneuvers, is in horizontal flight 35 m above the level ground. Suddenly, the plane encounters terrain that slopes gently upward at 4.3^{°}, an amount difficult to detect; see below figure-1. How much time does the pilot have to make a correction if the plane is to avoid flying into the ground? The airspeed is 1300 km/h.

**
Solution:- **

**Given:**

Inclination of slope, *?* = 4.3^{°}.

Altitude of plane, *h* = 35 m.

Speed of air, *v*_{air} = 1300 km/h.

Let us assume that the length of slope to the point where the plane meets the slope is given by length *l*. Also, the speed of the plane is equal in magnitude with that of the speed of air, therefore the speed of the plane (say *v*_{plane}) is equal to 1300 km/h.

The figure below shows the motion of the plane and various other variables

From the figure one can see that the altitude (say *d*) of the slope is given in terms of length *l* of slope and the angle *?* as:

*d* = *l *sin *?*

If the plane were to collide with the slope then the following condition should be fulfilled:

*h *- *d* = 0

*d* = *h*

Substitute the value of *d* and *h* to have,

*l *sin *?* = 35 m

*l* = 35 m/ sin (4.3^{°})

So, *l* = 466.7 m

Therefore the length of the slope after which the plane collides is 466.7 m.

Now, to calculate the horizontal distance (say *x*) that the plane will travel if it collides with the slope, we calculate the base of triangle ABC as

From triangle ABC, we have

*x* = *l *cos *?*

Substitute the value of *l* and *?*, to have

*x *= (466.7 m) [cos (4.3^{°})]

= 465.3 m

Therefore the horizontal distance at which the plane will collide with the slope is 465.3 m.

The time (say *t*) that the plane had before it run into the inclined ground ahead is given as:

*t* = *x*/*v*_{plane}

Substituting the value of *x* and *v*_{plane} in the equation *t* = *x*/*v*_{plane}, we get,

*t* = (465.3 m)/(1300 km/h)

=(465.3 m)/[(1300 km/h) (10^{3} m/1 km) (1 h/3600 s)]

=(465.3 m)/(361.1 m/s)

= 1.28 s

Rounding off to two significant figures,

*t* = 1.3 s

Therefore the plane has 1.3 s before it will run into the ground, and avoid the collision.

**Question 3:-**A juggler juggles 5 balls with two hands. Each ball rises 2 meters above her hands. Approximately how many times per minute does each hand toss a ball?

**Concept:**

First calculate the time for which a ball stays in air. Suppose that the time taken by the ball to reach the juggler hand from height say *y* is *t*. Since the balls begin with initial speed say *v*_{0y} of zero, the time *t* can be calculated as:

*y* = *v*_{0y}*t* + ½ *gt*^{2}

*y *= ½ *gt*^{2} (for *v*_{0y} = 0 m/s) …… (1)

The convention chosen is such that the downward motion is consider as positive.

The time (*t*) calculated above accounts for the time taken by the ball to reach the hand of the juggler from the maximum height it attains, therefore to calculate the time say *t*^{'}, for which the ball stays in the air, multiple the time (*t*) by 2.

By the time the ball left from one hand and reached the other, the juggler must have thrown 5 balls so that he can be ready to throw the ball reaching his hand again.Therefore, if the juggler has thrown 5 balls in time* t*^{'}, he would throw (5/*t*^{'}) (1 min) times each ball with his two hands in 1 min.

Thus the number of times say *n*, the ball is tossed by each hand is given as:

*n* = (5/*t*^{'})(1 min)/2 …… (2)

**Solution:**

Given that the ball rise 2 m above the hand, the distance *y* travelled by the ball when he descends from the maximum height to reach the juggler hand is 2 m. Also the free fall acceleration is 9.81 m/s^{2}.

Substitute the given values in equation (1),

*y *= ½ *gt*^{2}

2 m = ½ (9.81 m/s^{2}) *t*^{2}

*t *= √4 m/(9.81 m/s^{2})

= 0.63 s

The time (*t *^{'}) for which a ball stays in the air is given by multiplying *t* by 2 as:

*t *^{'} = 2*t*

= 2 (0.63 s)

= 1.27 s

The juggler has to juggles each ball once before time *t *^{'} so that he gets ready to juggle the ball reaching him again.

Therefore if two hands toss 5 times in 1.27 seconds, then the number of times (*n*) each ball is tossed in 1 min is given using equation (2) as:

*n* = [(5 times/1.27 s) (1 min)]/2

= [(5 times/1.27 s) (60 s)]/2

= 234.9 times/2

= 117.4 times.

Round off to three significant figures,

*n* = 117 times.

Therefore the juggler juggles at a rate of 117 toss per minute with each hand.