# A projectile is shot from the ground at a velocity of 1 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?

Nov 30, 2017

0.198 seconds

#### Explanation:

The key part of this problem to understand is that it is only the vertical motion of the projectile which matters; assuming the ground is perfectly flat (and not sloped), the horizontal motion only affects how far away from you it lands, not how long it takes to land.

To find the initial vertical velocity, we use the sine trigonometric function according to this formula:

${v}_{y} = v \cdot \sin \theta$

Our initial velocity is $v = 1 \frac{m}{s}$, and the angle $\theta = \frac{7 \pi}{12}$. Substituting these values we find ${v}_{y}$:

${v}_{y} = \left(1 \frac{m}{s}\right) \sin \left(\frac{7 \pi}{12}\right) = 0.966 \frac{m}{s}$

As you can see, this vertical velocity is nearly the full initial velocity. This is reasonable, as $\frac{7 \pi}{12} = {105}^{o}$, which is almost straight up and down.

Now, there are 2 ways to proceed at this point. One way uses the common position formula $y = {y}_{i} + {v}_{y} t + \frac{1}{2} a {t}^{2}$ to solve for $t$ when $y = 0$, which requires solving a quadratic equation (can be tricky sometimes).

The other way is to exploit the symmetry of vertical motion under constant acceleration of gravity with no other factors (wind resistance, etc). We know the projectile is going upwards with an initial vertical velocity of ${v}_{y}$, and at some time $t$ the upward motion will stop, after which time the projectile will begin falling back downwards. This "stopping point" will happen exactly halfway between the starting point and the ending point when the projectile lands again. Thus, all we have to do is find out how long it takes for the projectile to stop moving upwards and double that time.

Use the common velocity formula, with $a = - 9.8 \frac{m}{s} ^ 2$, to find when the velocity $v = 0$:

$v = {v}_{y} + a t$

$0 \frac{m}{s} = 0.966 \frac{m}{s} + \left(- 9.8 \frac{m}{s} ^ 2\right) t$

$\left(9.8 \frac{m}{s} ^ 2\right) t = 0.966 \frac{m}{s}$

$t = \frac{0.966 m / s}{9.8 m / {s}^{2}} = 0.099 s$

It will take 0.099 seconds to stop rising, which means it'll take another 0.099 seconds to fall back down, for a total time in the air of 0.198 s.

Alternative

$y = {y}_{i} + {v}_{y} t + \frac{1}{2} a {t}^{2}$

$0 = 0 + 0.966 t + \frac{1}{2} \left(- 9.8\right) {t}^{2}$

$0 = 0.966 t - 4.9 {t}^{2}$

$4.9 {t}^{2} - 0.966 t = 0$

$t \left(4.9 t - 0.966\right) = 0$

$t = 0$

and

$4.9 t - 0.966 = 0 \implies 4.9 t = 0.966 \implies t = 0.197$