Question

Explain the concept of Projectile with its derivations !! hoping quality answer ..
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Answer 1

Projectile & Projectile Motion – definitions. When an object is in flight after being projected or thrown then that object is called a projectile and this motion under the influence of constant velocity along the horizontal and downward gravitational pull along the vertical is called Projectile Motion

Answer 2

$<b> <u> Heya! </b> </u>$

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★$<i> <u> Projectile Motion - Concept ★</i> </u>$

→ $<b> What is Projectile Motion ? </b>$

▪Whenever we throw a body into space , provided with some initial velocity (u) , and then the body travels there under the influence of gravity alone - such a projection is termed as a projectile motion .
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★ We have two cases for projection :-

$<b> <u>Case.1 Simple Projection </b> </u>$

$<u>→ Trajectory</u>$ : the path transcended by a projectile is termed as a projectile. The trajectory of a projectile is a parabola .

$<u>→ Time of flight</u>$ : the maximum time upto which a projectile remains in its motion is called time of flight .

$expression = > \sqrt{ \frac{2h}{g} }$

$<u>→ Horizontal Range</u>$ : the maximum horizontal distance covered by a projectile is called its Horizontal Range , represented by R.

R = u × t

Putting the value of t from the above equation we have

$expression = > u \sqrt{ \frac{2h}{g} }$

$<u>→ Velocity component</u>$ : it is represented by v .

$expression = \sqrt{u {}^{2} + g {}^{2} t {}^{2} }$

From this expression we get ,

$tan \beta = \frac{gt}{u}$

$<b> <u>Case.2 Angular Projection </b></u>$

=> For this case , the projection is made at some angle ∅ therefore we divide our components :

✔Cos ∅ along the horizontal

✔Sin ∅ along the vertical .

$<u>→ Trajectory </u>$ :it is again a parabola as in above case.

→$<u>Time of flight </u>$: the expression is a bit different in this case.

$expression = \frac{2u \: sin \alpha }{g}$

→$<u> Horizontal Range </u>$: again the expression is a bit different.

=> u cos ∅ × t

=> u cos ∅ × 2u sin∅/g

[ use the identity Sin 2∅ = 2 Sin∅Cos∅]

$expression = \frac{u {}^{2} }{g} sin2 \alpha$

=> For maximum Range , $<b>∅ =45°</b>$

Then Range = u²/g

→$<u> Maximum Height</u>$ Hmax. The maximum vertical height attained by a projectile at an angle .

$expression = \frac{u {}^{2} sin {}^{2} \alpha }{2g}$



$<i> Thanks !! </i>$
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