Question

A Knight of the Round Table fires off a vat of
burning pitch from his catapult at 13.7 m/s,
at40◦ above the horizontal.
The acceleration of gravityis9.8m/s2.
How long is it in the air?

Answer 1

The first step to solving this problem is to remember the distinction that horizontal and vertical motion are independent of one another. This question asks for how long the projectile is in the air, so we only care about vertical motion. Since the projectile is launched at an angle, we need to break up the vector into its vertical and horizontal components first.



Now that we've found the vertical velocity of this object, we simply need to break up the vertical motion into two separate components. First, the object will travel to a maximum height. Then the object will fall from the maximum height due to just the force of gravity.

Traveling to maximum height, we use the idea that the projectile is accelerating downwards, even though it is traveling upwards. Since it's upward velocity initially is 8.806 m/s, it will be immediately counteracted by the 9.8 m/s^2 due to gravity. The time for this to occur is simply the velocity divided by the acceleration.\[t=\frac{ Vy }{ a }=\frac{ 8.806 }{ 9.8 }=0.899s\]
Now we need to find the time taken to fall from the max height. Considering our motion has stopped, this is simply free fall due to gravity. However, we need to first find our max height, so that we know how far the projectile needs to fall by gravity. To find the max height, we will use the kinematic equation that relates displacement to initial velocity, time, and acceleration.\[v^{2}=u^{2}+2as\]Written for s:\[s=\frac{ v^{2}-u^{2} }{ 2a }=\frac{ (0)^2 - (8.806)^2}{ 2(-9.8) }=3.96m\]
Now that we have found the height our object is at when it stop traveling upwards, we can use the kinematic equation that relates displacement, acceleration, initial velocity, and time.\[s=ut+\frac{ 1 }{ 2 }at^2\]Written for t (remembering now u = 0): \[t=\sqrt{\frac{ 2s }{ a }}=\sqrt{\frac{ 2(3.96) }{ 9.8 }}=0.899\]
This is not always the case, but here, both values for time were actually different, but the same when rounded to 3 places. This shocked me at first, but the math is correct. Just don't assume this is always what you'll get, but it is possible.

So, to finish this problem, simply sum the times to get the total flight time:\[t=t1+t2=(0.899)+(0.899)=1.8s\]

Answer 2

Ugh it cut out the drawing I made haha, this should come before the explanation above: