IIT JEE , AIEEE Forum - Physics , Mechanics - Kinematics problem-3(from JEE)

iitkgp_bipin

Let OB be the x-axis and OA be the y-axis.

a_x = -gcos30 = (-g)(

3)/2
a_y = -gsin30 = g/2

Initially : u_x = 10

3 and u_y = 0
Finally : v_x = 0

v_x = u_x + a_xt
0 = 10

3 + {(-g)(

3)/2}.t
t = 2 sec...........................(b)

Velocity with which it strikes plane OB :
v_y = u_y + a_y.t = 0 + (-gsin30)(2) = -10 m/s (i.e. 10 m/s in negative y-direction).

Rest of the part is easy , for e) part find the coordinates of P and Q and apply distance formula.

I would discuss part d).

Now consider horizontal and vertical as axes.

Initially vertical velocity u_y = (10

3 )sin30 = 5

3

At max. height v_y = u_y - g.t = 0

t =

3/2

Hence vertical distance transversed = u_yt - (1/2)gt²

Hence total height attained = h + u_yt - (1/2)gt²

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