Question

A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Figure).

1. Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).
2. Time of flight.
3. β at which range will be maximum.

Answer 1

a. Since we have, `x = L`

`u_x = v_x = v_0 cos β`

`a_x = - g sin α`

And `t = T = (2v_0 sin β)/(g cos α)`

Hence, `s = u_xt + 1/2 a_xt^2`

Gives us, L = `v_0 cos B * T + 1/2 (- g sin α) T^2`

= `T[v_0 cos β - 1/2 g sin α T]`

Putting `T = (2v_0 sin β)/(g cos α)` given us,

⇒ `L = (2v_0^2 sin  β)/(g cos^2 α) cos(α + β)`

 That is the required range.

b. Let us first calculate the time of flight.

We have,

⇒ `s = ut + 1/2  gt^2`

Where, `s = 0`

`u = u_y = v_0 sin β`

`g  = g_y = - g cos α, t = T`

Hence, ⇒ `0 = v_0 sin β (T) + 1/2 (-g cos α)T^2`

i.e. ⇒ `T[v_0 sin β - T  g/2 cos α]` = 0

So either T is zero or `v_0 sin β - T  g/2 cos α`

That gives the time of flight as `T = (2v_0 sin β)/(g cos α)`

c. Let us consider the equation `Z = sin β cos(α + β)`

We can write `Z = sin β cos(α + β) = 1/2 [sin(2β + α) - sin α]`

Now for Z to be maximum, it must have `sin(2β + α) = 1`

i.e. `sin(2β + α) = sin 90^circ`

That gives, `2β + α = 90^circ`

Hence, `β = (90^circ - α)/2` or `β = pi/4 - α/2`