an arrow is shot into air with a velocity of 60 m/s at an angle of 30�

1.	an arrow is shot into air with a velocity of 60 m/s at an angle of 30°. find the maximum height the arrow at ends and the range of the arrow. please solve this problem with explanation...
Answer» 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 : $\sf:\implies\:R=\dfrac{u^2sin2\theta}{g}$ $\sf:\implies\:R=\dfrac{(60)^2sin2(30)^{\circ}}{10}$ $\sf:\implies\:R=360\times sin60^{\circ}$ $\sf:\implies\:R=360\times\dfrac{\sqrt3}{2}$ $\sf:\implies\:R=180\times\sqrt3$ $:\implies\:\underline{\boxed{\bf{\gray{R=311.4\:m}}}}$

an arrow is shot into air with a velocity of 60 m/s at an angle of 30°. find the maximum height the arrow at ends and the range of the arrow. please solve this problem with explanation...

Answer»

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 :

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