Question

a projectile is thrown with the velocity of 20 metre per second at an angle of 60 degree with the horizontal calculate its time of flight horizontal range and maximum height ​

Answer 1

Explanation:

taking g=10m/sec

then t=2u/g

therefore t=2×20/10

t=4 second.

Horizontal Range(R) =400×0.8/10

R=40×0.8

R=32 metre.

Maximum Height(H)= 400×0.6/2×10

H= 20×0.6

H=120 metre.

Hope u understood

Answer 2

Answer:

A projectile is thrown with the velocity of 20 metres per second at an angle of 60 degrees with the horizontal.

a)  Time of flight              =       3.5347 sec

b)  Horizontal range        =       23.6984 m

c)  Maximum height        =        15.3061 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    = 20 m/s

θ    =  $60^{0}$

g    = 9.8 m/s

The horizontal range is given by,

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

   $=\frac{\mathrm{20}^{2} \sin( 2 \times 60)}{\mathrm{9.8}}$

R   =  23.6984 m

Time of flight can be calculated as follows,

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

$\mathrm{T}=\frac{2\times \mathrm{20} \times\sin 60}{\mathrm{9.8}}$

( sin 60 = $\frac{\sqrt{3} }{2}$ )

T =  3.5347 sec

The maximum height is,

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

   $=\frac{\mathrm{20}^{2} \sin ^{2} \ 60}{2\times \mathrm{~9.8}}$

H  = 15.3061 m