When a child standing on a horizontal path throws a ball, it leaves her hand from a point that is 90cm vertically above the path. The ball clears a 4.5 m high wall that is 10.5 m away from where it was thrown.
Show that the least velocity required for this to occur is the same as the velocity acquired by a body falling 7.35m under gravity.
My solution :
Case1 : 7.35m falling body
$$ v^2=u^2+2as \\ v^2=0+2(-9.8)(-7.35) \\ v=\sqrt{144.06} \\ v=12 \, \mbox{m}/\mbox{s} $$
Case 2: projectile motion
$$ t=0 \Rightarrow \\ \mathbf{V} = u \cos \alpha \, \mathbf{i} + u \sin \alpha \, \mathbf{j} \\ \mathbf{r} = 0.9 \, \mathbf{j} $$
$$ \mathbf{V} = u \cos \alpha \, \mathbf{i} + [ u \sin\alpha -9.8t] \, \mathbf{j} \\ \mathbf{r}= (u\cos \alpha t) \, \mathbf{i} + (0.9+ u \sin \alpha t-4.9 t^2) \, \mathbf{j} $$
I end up with the equations:
usina-9.8t=0
u=(9.8t)/sina......1
10.5=ucosat......2
4.50.9+usinat-4.9t^2......3
wrong answer where i let a=45
let angle a=45
0.9+usin(45)t-4.9t^2=4.5_________1
ucos(45)t=10.5 => y=10.5/[cos(45)*t]
sub 2 to 1
-4.9t^2+10.5/[cos(45)*t]*sin(45)*t-3.6=0
-4.9t^2+10.5tan(45)-3.6=0
-4.9t^2=-6.9
t=1.1867
so y=10.5/[cos(45)*1.1867] =12.51
wrong answer^
which is... not = to 12
$v = u \cos(α)\vec{i} + u \sin(α)\vec{j}$
Now, for $v$ to be valid, either $\frac{3.6}{u\sin(α)} = \frac{10.5}{u\cos(α)}$ which gives $α=18.92$
Or, $u\sin(α)t - 0.5 g t^2 = 3.6$ and $t=\frac{10.5}{u\cos(α)}$
Using these two equations, you can obtain the minimum $v$ required. The basic concept is that if, time taken to reach the wall is $t$, $t \times v_{\tiny \mbox{horiz}} = 10.5$ and $v_{\tiny \mbox{vert}} \times t - 0.5 g t^2 = 3.6$