# A projectile is shot at an angle of pi/4  and a velocity of  2 m/s. How far away will the projectile land?

Feb 4, 2016

I got $\text{Distance" ~=0.408" metres}$
To give a full explanation of the calculation method proved too long. So I have given a summery of the method I used.

#### Explanation:

$\textcolor{b l u e}{\text{Target: Time of flight x horizontal component of projectile}}$

Let total time of flight be ${T}_{f}$

Find time of flight using

Vertical component of projectile$= 2 \times \sin \left(\frac{\pi}{4}\right) \to \frac{2}{\sqrt{2}}$
The vertical travel distance (upwards) at ${T}_{f} \to \frac{2}{\sqrt{2}} \times {T}_{f}$

This must match the downward distance due to gravity at$\text{ } {T}_{f}$.

This is the mean velocity (downwards) multiplied by flight time

$\text{ } \frac{9.81 \times {T}_{f}}{2} \times {T}_{f}$

Equate them to each other and you have a quadratic that can be solved using the standard form equation.

$\text{ I got } {T}_{f} \cong 0.2883$ seconds
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Having found ${T}_{f}$ you then substitute into the distance equation of the horizontal component of the projectile.

$2 \times \cos \left(\frac{\pi}{4}\right) \times {T}_{f} \text{ " ->" } \frac{2 {T}_{f}}{\sqrt{2}}$

$\text{ Distance" ~=0.408" metres}$

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$\textcolor{b l u e}{\text{Comment}}$

$\textcolor{b r o w n}{\text{The "underline("actual")" height at any time "T_i" where } 0 \le {T}_{i} \le {T}_{f}}$ is

$\left(\frac{2}{\sqrt{2}} \times {T}_{i}\right) - \left(\frac{9.81 \times {T}_{i}^{2}}{2}\right)$