# Question #fefc4

Apr 12, 2017

No, it does not go over the wall

#### Explanation:

The key is to find the time of flight and use the idea that this is common to both horizontal and vertical components of motion.

The horizontal component of the velocity is given by:

$\textsf{{V}_{x} = v \cos \theta = \frac{45}{t}}$

$\therefore$$\textsf{t = \frac{45}{v \cos \theta} = \frac{45}{22 \cos 45} = 2.89 \text{ } s}$

Now we consider the vertical component:

We can use:

$\textsf{s = u t + \frac{1}{2} a {t}^{2}}$

Using the convention "up is +ve" this becomes:

$\textsf{s = v \sin \theta t - \frac{1}{2} {\text{gt}}^{2}}$

$\therefore$$\textsf{s = 22 \sin 45 \times 2.89 - \frac{1}{2} \times 9.81 \times {2.89}^{2}}$

$\textsf{s = 44.96 - 40.96 = 4.0 \textcolor{w h i t e}{x} m}$

You can see that it will not make it over the wall.