08H-37 Projectile Problem

[moy2414]

A projectile is fired at 50ft/s and hits a target 60ft away. The firing angle may be either less than 45° or greater than 45°. What is the positive difference in time of flight for these two paths?

Answer: 1.48 seconds

How do I work this problem?

[TFugere]

Since the velocity is at an angle, it can be broken into vcos0 and vsin0. Also, y=vsin0t+1/2at^2. When the horizontal equation D=Vcos0t is solved for t (D/Vcos0), it can be inserted into the y equation to get y=Dtan0 + (gD^2)/(2(Vcos0)^2). "g" is equal to 32.185 ft/s^2. Inserting this into the solver in a Ti89 or Ti84 and inserted 0 for y and your variable x as the angle "0," the two angles are 25.29 and 62.81 degrees. The time of flight is equal to 2Vsin0/g, and this is in ft/s, so the equation would be 2(50)/32 * (sin62.81 - sin25.29) which gives you 1.49 seconds, which is close enough to the actual answer.

[RussiaPSJA]

Just offering my two cents. If you prefer, you could use TFugere's method, but this is how I approach the question.

Looking at the question, we know how far the object goes horizontally, 60 feet, and we know the velocity: 50 ft/s. Also, when dealing with these questions, you need to know that the acceleration of gravity in ft/s^2 is 32.17 (Something you should have memorized for calculator).

So, right from the get go I'm thinking about the Distance Horizontal Max formula (Dhmax), which is. D=(V^2*Sin(2*theta))/G. I'm going to plug in the respected values into the equation wherever they belong and solve for theta:

60 feet = ([50 ft/s]^2*Sin(2*theta))/32.17 ft/s^2

Depending on how your solver works and what calculator your using it might work differently for getting both angles. As for my HP-35S:

Something to note is that you need two angles, so I'm going to use my solver TWICE, once for the angle below 45, and one for the angle above 45. I do this by storing a value below 45 in the theta variable I use the first time, and a value above 45 for the theta variable I use the second time. After solving for theta both times I arrive at theta = 25.27 degrees and 64.729 degrees.

Now, the time the object is in the air is independent of its horizontal distance, because gravity is a downward force, so only the vertical distance is important in the calculation of how long the object stays in the air for. We're going to utilize 2 formulas here, one to find the time, and the other to find it's maximum vertical distance. We could use the Distance Vertical Max formula (Dvmax) [D = (V^2*Sin(theta)^2)/(2*G)] for the distance, but the calculator test is a rigorous timed test, so we're going to use another formula, which works equally as well (but quicker) to find the Maximum Vertical Distance given the Maximum Horizontal distance and the corresponding angle. That formula is: Tan(theta) = 4*Dvmax/Dhmax.

So, solving for Dvmax:

Dvmax = Tan(theta)*Dhmax/4.

So, our Dvmax comes out to:

Tan(25.27)*60/4 = 7.081 feet, and Tan(64.729)*60/4 = 31.774 feet.

Now, we're going to finally find the time using the formula D= 1/2*a*t^2, where a is acceleration due to gravity in this case and t is time. The key thing to remember when using this formula is that it will only get the time for the projectile only reaching the height (or more technically, dropping from the height to the ground), so we're going to have to multiply the time by 2 to get the amount of time it takes for the projectile to get from the ground to the height, and then back down to the ground.

Solving for t:

7.081 feet = 1/2 * 32.17 * t^2; t = .6635 seconds. .6635 seconds * 2 = 1.326 seconds.
31.774 feet = 1/2 * 32.17 * t^2; t = 1.405 seconds. 1.405 seconds * 2 = 2.811 seconds.

Now, without further ado we simply subtract our times because that is what the question asks for:

2.811 - 1.326 = 1.48

It's just my two cents, and another way of doing the question, although its similar to TFugere's method.

[AuSmith]

So, here's my nonsolution

If you look at the formula

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you'll see that a projectile goes just as far when you shoot at an angle [unparseable or potentially dangerous latex formula] as when you shoot at [unparseable or potentially dangerous latex formula]. The time it takes to get to the top is [unparseable or potentially dangerous latex formula], and so the total time is [unparseable or potentially dangerous latex formula]. Similarly, it takes [unparseable or potentially dangerous latex formula] if you shoot at the other angle. Subtracting and using the fact that [unparseable or potentially dangerous latex formula],

time = [unparseable or potentially dangerous latex formula]