A projectile is fired horizontally off a cliff at an initial speed of `V` metres per second.
The projectile strikes the water, `l` metres from the base of the cliff.
Let `g` be the acceleration due to gravity and assume air resistance is negligible.
- Show the projectile hits the water when
`qquadt = sqrt((2d)/g)` (2 marks)
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- If `l` equals twice the height of the cliff, at what angle does the projectile hit the water? (2 marks)
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- Show that the speed at which the projectile hits the water is
`qquad2sqrt(dg)` metres per second. (1 mark)
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Show Worked Solution
| i. | `underset~v` | `= Vcosthetaunderset~i + (Vsintheta – g t)underset~j` |
| `= Vcos0 underset~i + (Vsin0 – g t)underset~j` | ||
| `= Vunderset~i – g tunderset~j` |
| `underset~s` | `= intunderset~v\ dt` |
| `= Vtunderset~i – 1/2g t^2 underset~j + c` |
`text(When)\ t = 0, underset~s = 0, c = 0`
`text(Time of flight:)`
`underset~j\ text(component of)\ underset~s = −d`
| `−1/2 g t^2` | `= −d` |
| `t^2` | `= (2d)/g` |
| `t` | `= sqrt((2d)/g)` |
ii. `l = 2d\ \ (text(given))`
`text(Projectile hits water at)\ theta:`
| `overset.y` | `= underset~j\ text(component of)\ underset~v` |
| `= −g t` | |
| `= −g · sqrt((2d)/g)` | |
| `= −sqrt(2dg)` | |
| `overset.x` | `= underset~i\ text(component of)\ underset~v` |
| `= V` |
`text(When)\ \ t = sqrt((2d)/g),`
`underset~i\ text(component of)\ underset~s = 2d`
| `2d` | `= V · sqrt((2d)/g)` |
| `V` | `= (2dsqrtg)/sqrt(2d) = sqrt(2dg)` |
| `tantheta` | `= (|overset.y|)/(|overset.x|)= sqrt(2dg)/sqrt(2dg)=1` |
| `:. theta` | `= 45°` |
iii. `text(Speed = magnitude of velocity)`
| `|underset~v|` | `= sqrt((sqrt(2dg))^2 + (sqrt(2dg))^2)` |
| `= sqrt(4dg)` | |
| `= 2sqrt(dg)` |