# Projectile motion question

• Mar 14th 2010, 06:13 PM
koopatroopa
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?
• Mar 14th 2010, 08:57 PM
sa-ri-ga-ma
In a projectile motion, to reach a target, the velocity of projection will be minimum when the angle of projection is 45 degrees.
• Mar 15th 2010, 03:00 AM
Hello koopatroopa

Welcome to Math Help Forum!
Quote:

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.)