A stone is thrown from the top of a tower 50m high with speed 30m/s at an angle 30° with horizontal. Find the distance from the foot of tower where the stone will hit the ground.

Here,       u = 30 m/s,       straight theta equals 30 degree
∴             straight u subscript straight x equals 30 space cos space 30 degree space equals space 25.98 space straight m divided by straight s
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#6 {main}</pre>
    Let the stone reaches the ground after time t.
                space space space 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

Here,       u = 30 m/s,       ∴             and    
Here,      straight y equals 0 comma space space straight y subscript straight o equals 50 comma space space space straight a subscript straight y equals negative 10 straight m divided by straight s squared
               0 equals 50 plus 15 straight t minus 5 straight t squared
or            straight t equals 5 straight s
    The distance from the foot of tower where the stone will hit the ground is,
              s = u x t = 25.98 x 5 = 104.9m

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Define relative velocity.


Relative velocity of a body A w.r.t. B is the velocity with which A appears to move to B, if B were at rest.

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#6 {main}</pre> is the velocity of B, then relative velocity of A w.r.t. B is, 

                       space space stack straight v subscript straight r with rightwards arrow on top space equals space stack straight v subscript straight A with rightwards arrow on top minus stack straight v subscript straight B with rightwards arrow on top

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Show that if the projectile attains the maximum height of H and falls at a distance R, then it is projected at an angle θ given by <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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#6 {main}</pre>

Let the speed with which the projectile be projected at angle θ  be 'u'.

The maximum height and the horizontal range is given by,

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

Maximum height, straight H equals fraction numerator straight u squared sin squared straight theta over denominator 2 straight g end fraction 
Ratio of horizontal range to maximum height is given by, 

∴               straight R over straight H equals fraction numerator 2 s i n 2 straight theta over denominator s i n squared straight theta end fraction equals 4 space c o t space straight theta

rightwards double arrow          t a n space straight theta space equals space fraction numerator 4 straight H over denominator straight R end fraction

That is, 
                  straight theta equals t a n to the power of negative 1 end exponent open parentheses fraction numerator 4 straight H over denominator straight R end fraction close parentheses, is the angle of projection. 
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Two bodies are placed at a distance 4 km and 6·5 km from the gun. When the bullet is fired at an angle of 30° with horizontal, it hits the body at 4 km. At what angle the bullet should be fired so that it may hit the second body?


Horizontal range is given by, 

                       straight R equals fraction numerator straight u squared sin space 2 straight theta over denominator straight g end fraction

Therefore the ratio of horizontal ranges R1 and R2 for angles of projections θ1 and θ2 is,

                 space space straight R subscript 1 over straight R subscript 2 equals fraction numerator sin space 2 straight theta subscript 1 over denominator sin space 2 straight theta subscript 2 end fraction

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#6 {main}</pre>

We have here,

Angle space of space projection comma space straight theta subscript 1 equals 30 degree

Horizontal space range comma space straight R subscript 1 space equals space 4 space km

Horizontal space range comma space straight R subscript 2 equals 6.5 space km

Therefore, 

sin 2 straight theta subscript 2 equals sin 60 degree cross times fraction numerator 6.5 over denominator 4 end fraction equals 1.407 greater than 1, which is not possible.

Therefore for same speed of bullet, the bullet can not hit the body at a distance of 6·5km.

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A body is projected with a velocity 30 m/sec at angle 30°. After what time will it be at maximum height?

We know,

A projectile is at maximum height at time equal to half of the time of flight.

That is, 

                space space straight t equals fraction numerator straight u space s i n space straight theta over denominator straight g end fraction

Here, we have

Initial space velcoity comma space straight u space equals space 30 space straight m divided by straight s

Angle space of space projection comma space straight theta space equals space 30 degree

Time when the projectile is at maximum height is, 

∴           straight t equals fraction numerator 30 cross times s i n space 30 degree over denominator 9.8 end fraction equals 1.53 straight s 
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