1. ## Projectile motion question

I'm having trouble with a Yr.12 Maths C question on projectile motion (I hope this is the right section to post it)

The question is: when a child standing on a horizontal path throws a ball, it leaves her hand from a point that is 90cm (0.9m) vertically above the path. The ball clears a 4.5-m-high wall that is 10.5m away from where it was thrown. SHow that "the least" velocity required for this to occur is the same as that acquired by a body falling 7.35m under gravity.

I've tried this a few different ways- but keep getting stuck with multiple variables, etc.
Can anyone help me out with this?

my attempt, (note: X=theta)
horizontal range= vcos(X)t and displacement(vertical)= vsin(X)t-4.9t^2 +0.9
if x=10.5 and Y=3.6 at one point- we can sub these into the equations to get:

Substitute and rearrange for v
10.5= vcos(X)t {v=10.5/(cos(x)t)}
4.5= vsin(x)t-4.9t^2+0.9 {v=(4.5+4.9t^2-0.9)/sin(x)t}

therefore,
10.5/(cos(x)t)=v=(4.5+4.9t^2-0.9)/sin(x)t
simplifying,
4.9t^2=10.5tan(X)-3.6

this equation has 2 variables (the angle and time), is there any way to solve this {i recall my teacher mentioning it- but not in great detail}. Or is this the wrong way to attempt it?

2. In a projectile motion, to reach a target, the velocity of projection will be minimum when the angle of projection is 45 degrees.

3. Hello koopatroopa

Welcome to Math Help Forum!
Originally Posted by koopatroopa
I'm having trouble with a Yr.12 Maths C question on projectile motion (I hope this is the right section to post it)

The question is: when a child standing on a horizontal path throws a ball, it leaves her hand from a point that is 90cm (0.9m) vertically above the path. The ball clears a 4.5-m-high wall that is 10.5m away from where it was thrown. SHow that "the least" velocity required for this to occur is the same as that acquired by a body falling 7.35m under gravity.

I've tried this a few different ways- but keep getting stuck with multiple variables, etc.
Can anyone help me out with this?

my attempt, (note: X=theta)
horizontal range= vcos(X)t and displacement(vertical)= vsin(X)t-4.9t^2 +0.9
if x=10.5 and Y=3.6 at one point- we can sub these into the equations to get:

Substitute and rearrange for v
10.5= vcos(X)t {v=10.5/(cos(x)t)}
4.5= vsin(x)t-4.9t^2+0.9 {v=(4.5+4.9t^2-0.9)/sin(x)t}

therefore,
10.5/(cos(x)t)=v=(4.5+4.9t^2-0.9)/sin(x)t
simplifying,
4.9t^2=10.5tan(X)-3.6

this equation has 2 variables (the angle and time), is there any way to solve this {i recall my teacher mentioning it- but not in great detail}. Or is this the wrong way to attempt it?
You've started off with the correct equations, but you need to eliminate $t$ between them to get:
$4.5=10.5\tan\theta -\frac{4.9\cdot10.5^2\sec^2\theta}{v^2}+0.9$

$\Rightarrow 3.6v^2=10.5v^2\tan\theta -4.9\cdot10.5^2(1+\tan^2\theta)$

$\Rightarrow 4.9\cdot10.5^2\tan^2\theta -10.5v^2\tan\theta +3.6v^2+4.9\cdot10.5^2=0$

Now here's the tricky bit. Thinking of this as a quadratic in $\tan\theta$, the minimum value of $v$ occurs when this equation has equal roots. This corresponds, of course, to there being just one value of $\theta$ for the ball to just clear the wall.

So you now set the discriminant equal to zero to get:
$(10.5v^2)^2 - 4\cdot4.9\cdot10.5^2(3.6v^2+4.9\cdot10.5^2)=0$

$\Rightarrow(v^2)^2 - 4\cdot4.9\cdot3.6v^2-4\cdot4.9\cdot4.9\cdot10.5^2=0$, after dividing through by $10.5^2$
Using the quadratic formula on this equation, taking the positive root (since $v^2 > 0$) gives
$v^2 = 144.06$
Then, using $v^2 = 2gh$:
When $g = 9.8$ and h = $7.35$
$v^2 = 144.06$
and hence the required result.

(Note: the angle of $45^o$ only applies to the maximum range on a horizontal plane; and, of course, the top of the wall is not at the same height as the point of projection.)