A projectile is shot from the ground at a velocity of 19 ms^-1 and at an angle of (2pi)/3. How long will it take for the projectile to land?

Jan 20, 2016

Initial velocity in the y direction, ${v}_{y} = 16.5 m {s}^{-} 1$. This falls to $0$ in $1.68 s$ at the top of the trajectory, so the full trajectory back to earth takes $3.36 s$.

Explanation:

The approach to solving projectile motion problems is to divide the motion into its horizontal component, ${v}_{x}$, and its vertical component, ${v}_{y}$.

${v}_{x} = v \cos \theta = 19 \cos \left(\frac{2 \pi}{3}\right) = - 9.5 m {s}^{-} 1$ (since $\cos \left(\frac{2 \pi}{3}\right) = - .0 .5$

This just means the initial velocity in the $x$ direction is to the left. If we ignore air resistance, ${v}_{x}$ will be constant throughout the projectile's flight. For this problem we won't use this value again.

${v}_{y} = v \sin \theta = 19 \sin \left(\frac{2 \pi}{3}\right) = 16.5 m {s}^{-} 1$

${v}_{y}$ will start out at its maximum value, decrease to $0 m {s}^{-} 1$ under the influence of gravity as the object reaches the highest point of its trajectory, then increase again to return to its initial value as the projectile returns to ground level.

To solve this particular problem, we divide the trajectory into two equal parts: before the projectile reaches its maximum height and after. These two parts will take the same amount of time.

We use the fact that the velocity in the y direction falls to $0$ at the top of the trajectory:

$v = u + a t$

Rearranging to make $t$ the subject:

$t = \frac{v - u}{a} = \frac{0 - 16.5}{-} 9.8 = 1.68 s$

(the acceleration due to gravity is shown as negative because it is in the opposite direction to the initial velocity)

Crucial to remember - students often forget this in the exam - that this is the time for only half the trajectory: the object still needs to return to earth. Double this time and the final answer is $3.36 s$.