Question

a stone is thrown from the horizontal ground at an angle of projection of 30° with a speed of 49 m/s. Calculate it's (1)time of flight (2) horizontal range.​

Answer 1

Answer:

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Explanation:

a stone is thrown from the horizontal ground at an angle of projection of 30° with a speed of 49 m/s. Calculate it's (1)time of flight (2) horizontal range.

Answer 2

Answer:

A stone is thrown from the horizontal ground at an angle of projection of 30° with a speed of 49 m/s.  

( 1 ) Time of flight          =  5 sec

( 2 ) horizontal range   =  200 m

Explanation:

For a projectile motion,

• Time of flight        :       $\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$  
• Horizontal range  :      $\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$

• Maximum height  :      $\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$

where,

u - Initial velocity

θ - Angle of projection

g - Acceleration due to gravity

Given,

u    = 49 m/s

θ    =  $30^{0}$

g    = 9.8 m/s

Time of flight can be calculated as follows,

$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$

$\mathrm{T}=\frac{2\times \mathrm{49} \times\sin 30}{\mathrm{9.8}}$

( sin 30 = $\frac{1 }{2}$ )

T =  5 sec

The horizontal range is given by,

$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$

   $=\frac{\mathrm{49}^{2} \sin( 2 \times 30)}{\mathrm{9.8}}$

R   = 200 m