Question

a body is projected with a velocity of 20 metre per second in the direction making an angle of 60 degree what is the horizontal calculate its position after 0.5 second and velocity after zero​

Answer 1

Answer:

Explanation:

Here, u = 20 m/s, θ = 600

Let, vx, vy be the rectangular components of velocity of body after 0.5 s of projection.

Where,

vx = u cos 600 = 20 X 1/2 = 10 m/s

Taking vertical upward motion of body for 0.5 s, we have :

vy = ? , uy = u sin 600 = 20 X 3√2 = 103√ ms−1

ay = -9.8 m/s2

t = 0.5 s

​As, vy = uy + ayt

​vy = 10 3√ + (−9.8) X 0.5  = 12.2 m/s

Resultant velocity,

v= −√vx^2 + vy^2

 = √  10^2 + 12.2^2 

= 15.77 ms^-1

Position of body after 0.5 s along the vertical direction is given by :

y = y0 + uy t + 1/2 ay t^2    .........(1)

​Here,

y0 = 0, uy = u sin 600, ay = -g

​Putting these values in eq. 1, we get :

y = 0 + u sin 600  t + 1/2 (-g) t^2

​y = u sin 600 t - 1/2 gt^2

​Putting,  

u = 20 m/s and t = 0.5 s in above equation, we get :

y = 7.32 m