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?