Question

plzzzzzz.... help...... ​

Answer 1

Given :-

Initial velocity = 60 m/s

Angle of inclination = 30°

To Find :-

Maximum height reached and range of projectile.

Solution :-

❒ A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration is equal to g.

A] Maximum Height :

$\sf:\implies\:H=\dfrac{u^2sin^2\theta}{2g}$

$\sf:\implies\:H=\dfrac{(60)^2\times sin^230^{\circ}}{2(10)}$

$\sf:\implies\:H=\dfrac{3600\times (1/2)^2}{20}$

$\sf:\implies\:H=180\times \dfrac{1}{4}$

$:\implies\:\underline{\boxed{\bf{\purple{H=45\:m}}}}$

B] Horizontal Range :

$\sf:\implies\:R=\dfrac{u^2sin2\theta}{g}$

$\sf:\implies\:R=\dfrac{(60)^2sin2(30)^{\circ}}{10}$

$\sf:\implies\:R=360\times sin60^{\circ}$

$\sf:\implies\:R=360\times\dfrac{\sqrt3}{2}$

$\sf:\implies\:R=180\times\sqrt3$

$:\implies\:\underline{\boxed{\bf{\gray{R=311.4\:m}}}}$

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Answer 2

Answer:

Solution :-

A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration is equal to g.

A] Maximum Height :

$\sf:\implies\:H=\dfrac{u^2sin^2\theta}{2g}$

$\sf:\implies\:H=\dfrac{(60)^2\times sin^230^{\circ}}{2(10)}$

$\sf:\implies\:H=\dfrac{3600\times (1/2)^2}{20}$

$\sf:\implies\:H=180\times \dfrac{1}{4}$

$:\implies\:\underline{\boxed{\bf{\blue{H=45\:m}}}}$

B] Horizontal Range :

$\sf:\implies\:R=\dfrac{u^2sin2\theta}{g}$

$\sf:\implies\:R=\dfrac{(60)^2sin2(30)^{\circ}}{10}$

$\sf:\implies\:R=360\times sin60^{\circ}$

$\sf:\implies\:R=360\times\dfrac{\sqrt3}{2}$

$\sf:\implies\:R=180\times\sqrt3$

$:\implies\:\underline{\boxed{\bf{\blue{R=311.4\:m}}}}$

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