smc-1087-50-Angle at Impact

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 `t = 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 `2sqrt(dg)` metres per second. (3 mark)

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a. `text(See Worked Solutions)`

b. `45°`

c. `text(See Worked Solutions)`

Show Worked Solution

a.	`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`

b. `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°`

c. `text(Speed = magnitude of velocity)`