A projectile is projected with velocity u at an angle θ with horizontal. Discuss the trajectory of projectile. Also find the velocity of projectile at any instant.

Let us consider a projectile is projected with velocity u at angle θ with horizontal direction from a point on the ground. Let the point of projection be origin, right direction is positive direction of X-axis and vertically upward direction is positive direction of Y-axis.

Equation of trajectory:
Break the components of velocity and acceleration due to gravity in horizontal and vertical direction.

Let us consider a projectile is projected with velocity u at angle θ

Motion along horizontal direction:
straight x equals straight x subscript straight o plus straight u subscript straight x straight t plus 1 half straight a subscript straight x straight t squared equals straight u space cosθt            ...(1)
Motion along vertical direction:
space space space space straight y equals straight y subscript straight o plus straight u subscript straight y straight t plus 1 half straight a subscript straight y straight t squared
        equals usinθt minus 1 half gt squared                             ...(2)
Eliminating t from (1) and (2),
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This the required equation of trajectory. Since the equation of trajectory is the equation of parabola, therefore projectile follows parabolic path.

Let us consider a projectile is projected with velocity u at angle θ
Velocity of projectile

Let us consider a projectile is projected with velocity u at angle θ

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A car starts from rest and accelerates uniformly with 2 m/s2. At t = 5s a stone is dropped out of the window 15 m high of the car. What is velocity and acceleration of stone at t = 5·5 s? ( g = 10m/s2).

Car starts from rest and accelerates with acceleration 2 m/s2. Therefore the velocity of car at t = 5s is 10 m/s. When stone is dropped the velocity of car and hence velocity of stone is 10 m/s in horizontal direction. Since at 5s the stone loses the contact from car, therefore, the stone will move uniformly in horizontal direction and accelerate in vertically downward direction due to gravity.
At t = 5·5s or 0·5 s after the stone is dropped, the X and Y component of velocity are
   
Car starts from rest and accelerates with acceleration 2 m/s2. Theref

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Two projectiles are projected with same speed but at different angles. Under what conditions can they land at the same point at the same time on the earth?

Given, two projectiles are projected at different angles with the same speed. 

So, the two angles of projection i.e. θ and 90°–θ, for which horizontal range is same.

The projectiles will land at the same point if they are projected at angles θ and (90°–θ) from the same point and in the same plane. But, since the time of flight is different for the two cases, therefore the projectiles will reach the ground at different moments. 

If the projectile that takes more time to reach the ground is projected earlier than the second, then both may reach the ground at the same moment. 

Time of flight of first projectile projected at angle θ, 

Time of flight of second projectile projected at angle 90°–θ,
 

If two projectiles are projected at the time interval, then they will land at the same time. 

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A projectile is projected with velocity u at an angle θ with horizontal. Find the time of flight, maximum height attained and horizontal range of projectile.


Let us consider a projectile is projected with velocity u at angle θ with the horizontal direction from a point on the ground. Let the point of projection be origin, right direction is positive

Let us consider a projectile is projected with velocity u at angle θ
direction of X-axis and vertically upward direction is positive direction of Y-axis. Break the components of velocity and acceleration due to gravity in horizontal and vertical direction.

Let us consider a projectile is projected with velocity u at angle θ
Time of flight: It is the time for which the projectile remains in air above the point of projection. It is represented by T.
The position of projectile at any instant in Y direction is given by,
space space space space straight y equals straight y subscript straight o plus straight u subscript straight y straight t plus 1 half straight a subscript straight y straight t squared
The time of flight is the difference of instants when the Y position is zero.
i.e.          0 equals straight y subscript straight o plus straight u subscript straight y straight t plus 1 half straight a subscript straight y straight t squared
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or                  space space straight t subscript 1 equals 0    and     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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Maximum height: It is the greatest height to which the projectile rises above point of projection. It is represented by H. The displacement of projectile in vertical direction is given by
straight S subscript straight x equals fraction numerator straight v subscript straight y squared minus straight u subscript straight y squared over denominator 2 straight a subscript straight x end fraction
At the highest point, vertical component of velocity is zero.
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Horizontal range: It is the distance covered by projectile in horizontal direction when it again arrives at the level of projection. It is represented by R. Obviously, it is the distance travelled by projectile in horizontal direction with constant velocity u cosθ in interval equal to time of flight.
i.e.      straight R equals ucosθT equals ucosθ fraction numerator 2 usinθ over denominator straight g end fraction
or        straight R equals fraction numerator straight u squared sin 2 straight theta over denominator straight g end fraction

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A player P kicks the football with velocity 20 m/s at angle of 15°. Another player Q on the goal line 12 m away in the direction of kick, runs at the same instant to meet the ball. With what speed he should run so that he could catch the ball just before it hits the ground? (Take g = 10 m/s2).

We know that,

Horizontal range, straight R equals fraction numerator straight u squared sin 2 straight theta over denominator straight g end fraction

We have,

Initial space velocity comma space straight u equals 20 straight m divided by straight s

Angle space of space projecvtion comma space straight theta equals 15 degree

a c c e l e r a t i o n space d u e space t o space g r a v i t y comma space straight g equals 10 space straight m divided by straight s squared

∴  

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#6 {main}</pre>

i.e. the ball hits the ground at a distance of 20m from the position of player P.

Since the player Q is 12m away, therefore, he has to travel a distance 8m to catch the ball in time equal to the time of flight.

Now the time of flight is,

        space space space space straight T equals fraction numerator 2 straight u space s i n theta over denominator straight g end fraction
space space space space space space equals fraction numerator 2 cross times 20 cross times s i n 15 degree over denominator 10 end fraction
space space space space space space space equals 4 s i n 15 degree

rightwards double arrow       straight T equals 1.035 space straight s 

The velocity of player Q is,

     straight v equals fraction numerator 8 over denominator 1.035 end fraction equals 7.727 space straight m divided by straight s

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