Question

a stone is thrown from the horizontal ground at an angle of projection of 30 degree with speed of 45 metre per second calculate the time of flight and horizontal range​

Answer 1

Answer:

R=175.75

Tf = 4.5

Explanation:

take g =10

and apply formula

1.R= u^2 sin2theta/g

2. Time of flight=2u sintheta/g

Answer 2

Answer:

A stone is thrown from the horizontal ground at an angle of projection of 30 degrees with a speed of 45 m/s.

Horizontal range    =  253.0722 m

Time of flight          =  0.510 sec

Maximum height    =  6.4572 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    = 45 m/s

θ    =  $30^{0}$

g    = 9.8 m/s

The horizontal range is given by,

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

   $=\frac{\mathrm{45}^{2} \sin( 2 \times 30)}{\mathrm{9.8}}$

R   = 253.0722 m

Time of flight can be calculated as follows,

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

$\mathrm{T}=\frac{2\times \mathrm{45} \times\sin 30}{\mathrm{9.8}}$

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

T =  0.510 sec

The maximum height is,

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

   $=\frac{\mathrm{45}^{2} \sin ^{2} \ 30}{2\times \mathrm{~9.8}}$

H  =  6.4572 m