Question

A body is projected with a velocity of 20m/s in a direction making an angle of 60˚ with the horizontal. calculate its i) position after 0.5 sec ii) velocity after 0.5 sec.

Answer 1

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

Answer 2

Answer:

Time of flight of the particle T=

2×10

2×20×

3

T=2

3

seconds

i)Position in X-direction after 0.55 second X=10×0.55=5.5m

In Y-direction ,Y=10

3

×0.55−5×(0.55)

2

Y=8m

position=(5.5,8)

ii)Velocity in X direction =10m/s

Velocity in Y direction v=u−gt

v=10

3

−10×0.55=11.8m/s

velocity =10

i

^

+11.8

j

^

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