Question

a projectile is thrown with a velocity 5metre per second at an angle of 30 degree with the horizontal. calculate range of projectile,time of flight and maximum height.​

Answer 1

Explanation:

the Area of a trapezium of height 12cm is 240cm square .if the length of one parallel side is 25cm . find the length of the other parallel side .

Answer 2

Answer:

A projectile is thrown with a velocity 5metre per second at an angle of 30 degrees with the horizontal.

a)   Range of projectile   =  3.1243 m

b)   Time of flight             =  0.51020 sec

c)    Maximum height      =​   0.3188 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:       $\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$

Given,

Initial velocity (u)            = 5 m/s

Angle of projection (θ)    =  $30^{0}$

Acceleration due to gravity (g) = 98 m/s

The horizontal range is given by,

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

   $=\frac{\mathrm{5}^{2} \sin( 2 \times 30)}{\mathrm{9.8}}$

  (sin 60 = $\frac{\sqrt{3} }{2}$  )

R   =  3.1243 m

Time of flight can be calculated as follows,

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

$\mathrm{T}=\frac{2\times \mathrm{5} \times\sin 30}{\mathrm{9.8}}$

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

T =  0.51020 sec

The maximum height is,

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

   $=\frac{\mathrm{5}^{2} \sin ^{2} \ 30}{2\times \mathrm{~9.8}}$

H  = 0.3188 m