Question

Projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The projectile hits the incline plane at point M.

a) Find the time it takes for the projectile to hit the incline plane.
b)Find the distance OM.​

Answer 1

Assume that point O is on the incline,
`g = 9.81\; {\rm m\cdot s^(-2)}`, and that air resistance is negligible.

Time taken: approximately
`0.65\; {\rm s}`.

Displacement along the slope: approximately
`9.1\; {\rm m}`.

Step-by-step explanation:

The time taken during the flight can be found in the following steps:

• Find an expression for the initial horizontal and vertical velocity (
  `u_(x)` and
  `u_(y)`) of this projectile.
• Let the duration of the flight
  `t` be an unknown.
• Find an expression for the vertical component of displacement,
  `x_(y)`, in terms of the duration of the flight,
  `t`.
• Find an expression for the horizontal component of displacement during the flight,
  `x_(x)`, in terms of the duration of the flight,
  `t`.
• Find an alternative expression for vertical displacement
  `x_(y)` from the expression for horizontal displacement
  `x_(x)` in the previous step.
• Equate the two expressions for vertical displacement
  `x_(y)` and solve for duration of the flight,
  `t`.

After the time taken has been found, the displacement along the slope can be found in the following steps:

• Using the expression for
  `x_(x)` in terms of initial horizontal velocity
  `u_(x)` and
  `t`, find the value of
  `x_(x)\!`.
• Divide horizontal displacement
  `x_(x)` by
  `\cos(15^(\circ))` (where
  `15^(\circ)` is the angle of elevation of the slope) to find the displacement
  `x` along the slope.

At
`22^(\circ)` above the horizontal direction, the horizontal and vertical components of initial velocity would be:

• Initial horizontal velocity:
  `u_(x) = (15\; {\rm m\cdot s^(-1)})\, \cos(22^(\circ)}) \approx 13.908\; {\rm m\cdot s^(-1)}`.
• Initial vertical velocity:
  `u_(y) = (15\; {\rm m\cdot s^(-1)})\, \sin(22^(\circ)}) \approx 5.6191\; {\rm m\cdot s^(-1)}`.

Assume that air resistance is negligible. If the duration of the flight is
`t`, vertical acceleration would be constantly
`a_(y) = -g = (-9.81)\; {\rm m\cdot s^(-2)}`. Vertical displacement during the entire flight would be:

`\displaystyle x_(y) = (1)/(2)\, (-g)\, t^(2) + u_(y)\, t`.

Also assuming that air resistance is negligible, horizontal velocity would be constant. Horizontal displacement during the flight would be:

`\displaystyle x_(x) = u_(x)\, t`.

Since the overall displacement
`x` is in the direction of the slope, at
`10^(\circ)` above the horizontal:

`\begin{aligned} x_(y) &= x_(x) \, \tan(10^(\circ)) \\ &= u_(x)\, t\, \tan(10^(\circ))\end{aligned}`.

Equate the two expressions for vertical displacement
`x_(y)` and solve for
`t`:

`\displaystyle (1)/(2)\, (-g)\, t^(2) + u_(y)\, t = u_(x)\, t\, \tan(10^(\circ))`.

`\begin{aligned}t &= (u_(x)\, \tan(10^(\circ)) - u_(y))/((1/2)\, g) \\ &\approx (13.908\, \tan(10^(\circ)) - 5.6191)/((1/2)\, (-9.81)) \; {\rm s}\\ &\approx 0.6456\; {\rm s}\end{aligned}`.

In other words, the duration of the flight would be approximately
`0.65\; {\rm s}`.

Also because the overall displacement is in the direction of the slope (
`10^(\circ)` above the horizontal,) the overall displacement can be found from horizontal displacement:

`\begin{aligned}x &= (x_(x))/(\cos(10^(\circ))) = (u_(x)\, t)/(\cos(10^(\circ)))\end{aligned}`.

Substitute
`u_(x) \approx 13.908\; {\rm m\cdot s^(-1)}` and
`t \approx 0.6456\; {\rm s}` into this expression to find the value of displacement
`x`:

`\begin{aligned}x &= (u_(x)\, t)/(\cos(10^(\circ))) \\ &\approx ((13.908)\, (0.6456))/(\cos(10^(\circ)))\; {\rm m} \\ &\approx 9.1\; {\rm m}\end{aligned}`.

In other words, the projectile would land approximately
`9.1\; {\rm m}` from where it was launched.