Question

A rocket is launched at 500 m/s 60° to the horizontal
A. Find the components of the velocity
B. How long will it take for the rocket to land
C. How high will the rocket fly
D. What is the range of the rocket

Answer 1

Given :

Initial velocity of rocket = 500 m/s

Angle of projection = 60°

To Find :

• Components of the velocity
• Time of flight
• Max. height
• Range

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] Components of the velocity :

• Horizontal component :

=> u cos θ

=> 500 × cos 60°

=> 500 × 1/2

=> 250 m/s

• Vertical component :

=> u sin θ

=> 500 × sin 60°

=> 500 × √3/2

=> 250√3 m/s

B] Time of flight :

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

$\sf:\implies\:T=\dfrac{2(500)\:sin(60^{\circ})}{10}$

$\sf:\implies\:T=100\times\dfrac{\sqrt3}{2}$

$:\implies\:\underline{\boxed{\bf{\purple{T=50\sqrt3\:sec}}}}$

C] Max. height :

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

$\sf:\implies\:H=\dfrac{(500)^2\:sin^2(60^{\circ})}{2(10)}$

$\sf:\implies\:H=\dfrac{250000\times (\sqrt3)^2}{20\times(2)^2}$

$\sf:\implies\:H=\dfrac{250000\times3}{80}$

$:\implies\:\underline{\boxed{\bf{\red{H=9375\:m}}}}$

D] Range :

$\sf:\implies\:R\:tan\theta=4H$

$\sf:\implies\:R\:tan(60^{\circ})=4(9375)$

$\sf:\implies\:R=\dfrac{37500}{\sqrt3}$

$:\implies\:\underline{\boxed{\bf{\gray{R=21676.3\:m}}}}$