Question

After one second the velocity of a projectile makes an angle of 45 with the horizontal after another second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are

Answer 1

Let a projectile is projected with speed u at an angle α with horizontal.

so, initial velocity , u = ucosα i + usinα j

and velocity after time t, v = ucosα i + (usinα - gt) j.....(1)

after one second, velocity of projection makes an angle 45° with horizontal.

so, velocity after 2s , v = ucosα i + (usinα - 2g) .[ from equation (1)]

but given, velocity of projectile after 2s, is only in horizontal direction.

so, v = ucosα i = ucosα i + (usinα - 2g) .

or, usinα - 2g = 0

or, usinα = 2g .......(2)

again, velocity after 1s, makes an angle 45° with horizontal.

i.e., V = v'cos45° i + v'sin45° j

so, velocity after one second, V = ucosα i + (usinα - g). [ from equation (1) ].

so, v'cos45° i + v'sin45° = ucosα i + (usinα - g) j

on comparing we get,

v'cos45° = ucosα

v' sin45° = (usinα - g)

from both equations we get,

ucosα = usinα - g ......(3)

from equations (2) and (3),

tanα = 2 => α = tan^-1(2)

and u = g/cosα = √5g

hence, angle of projection = tan^-1(2)

initial velocity = √5g