# A projectile is shot at an angle of pi/12  and a velocity of 4 m/s. How far away will the projectile land?

Mar 15, 2016

$s = 0.8 m$

#### Explanation:

Let the gravity acceleration be $g = 10 \frac{m}{s} ^ 2$

The time traveled will be equal to the time it reaches its maximum height ${t}_{1}$ plus the time it hits the ground ${t}_{2}$. These two times can be calculated from its vertical motion:

The initial vertical speed is:

u_y=u_0sinθ=4*sin(π/12)
${u}_{y} = 1.035 \frac{m}{s}$

Time to maximum height ${t}_{1}$

As the object decelerates:

$u = {u}_{y} - g \cdot {t}_{1}$

Since the object finally stops $u = 0$

$0 = 1.035 - 10 {t}_{1}$

${t}_{1} = \frac{1.035}{10}$

${t}_{1} = 0.1035 s$

Time to hit the ground ${t}_{2}$

The height during the rising time was:

$h = {u}_{y} \cdot {t}_{1} - \frac{1}{2} \cdot g \cdot {t}_{1}^{2}$

$h = 1.035 \cdot 0.1035 - \frac{1}{2} \cdot 10 \cdot {0.1035}^{2}$

$h = 0.05359 m$

The same height applies to the dropping time, but with the free fall formula:

$h = \frac{1}{2} \cdot g \cdot {t}_{2}^{2}$

${t}_{2} = \sqrt{\frac{2 h}{g}}$

${t}_{2} = 0.1035 s$

(Note: ${t}_{1} = {t}_{2}$ because of energy preservation law.)

The total time traveled is:

${t}_{t} = {t}_{1} + {t}_{2}$

${t}_{t} = 0.1035 + 0.1035$

${t}_{t} = 0.207 s$

The distance traveled in the horizontal plane has a constant speed equal to:

u_x=u_0cosθ=4*cos(π/12)
${u}_{x} = 3.864 \frac{m}{s}$

Finally, the distance is given:

${u}_{x} = \frac{s}{t}$

$s = {u}_{x} \cdot t$

$s = 3.864 \cdot 0.207$

$s = 0.8 m$

P.S. For future problems identical to this one but with different numbers, you can use the formula:

s=u_0^2*sin(2θ)/g

Proof : we are basically going to use the same method inversely, but without substituting the numbers:

$s = {u}_{x} \cdot {t}_{t}$

s=u_0cosθ*2t

s=u_0cosθ*2u_y/g

s=u_0cosθ*2(u_0sinθ)/g

s=u_0^2*(2sinθcosθ)*1/g

s=u_0^2*sin(2θ)*1/g

s=u_0^2*sin(2θ)/g