While sitting in a math class, you become curious and decide to put the math and your physics to work in a practical way. If you can shoot a spit-wad from a straw with a total speed of 6 m/s, at what angle would you have to shoot it in order for it to hit a blackboard that is 1 m from you, at a height that is even with your mouth? Keep in mind a trig identity: 2sin theta cos theta = sin2theta.
First off, why does it even have to be pointed at an angle (as opposed to straight ahead), and what is the approach to finding the angle once I've drawn it out, considering the velocity?
find time in air:
in the vertical direction:
hf=hi+6sinTheta*t-4.9t^2
but hf=hi=0, so t= 6sinTheta/4.9
in the horizontal direction:
distance=velocity*time
1m=6cosTheta*6sinTheta/4.9
Now you can use the angle formula you were given.
the article in wikipedia on Trajectory explains this well.
And of course there has to be an angle. It takes some time for the spit wad to get to the target, and it drops during that interval ...
When you shoot a spit-wad from a straw, it follows a projectile motion, meaning it moves in both the horizontal and vertical directions simultaneously. The reason you need to shoot it at an angle is because if you were to shoot it straight ahead, the spit-wad would only travel in the horizontal direction and would fall straight down due to gravity. By shooting it at an angle, you give it an initial vertical velocity component, allowing it to reach a higher height and potentially hit the blackboard.
To find the angle at which you need to shoot the spit-wad, you can break down the initial velocity into its horizontal and vertical components. Let's call the angle you're looking for theta. The horizontal component of velocity (Vx) is given by Vx = V * cos(theta), and the vertical component of velocity (Vy) is given by Vy = V * sin(theta), where V is the total speed of the spit-wad (6 m/s in this case).
Now, to calculate the angle that will allow the spit-wad to hit the blackboard, you need to consider the time and distance it takes for the spit-wad to travel. The time it takes for the spit-wad to reach the blackboard will be the same for both the horizontal and vertical motions. The distance traveled horizontally (range, R) is given by R = Vx * t, where t is the time of flight.
In this case, you are given that the blackboard is 1 m away from you horizontally. Hence, R = 1 m. Now, you need to find the time of flight (t). Since the spit-wad reaches its maximum height halfway through the flight, we know that the time taken to reach the maximum height will be half the total time of flight.
To find the total time of flight, you can use the vertical motion equation:
h = (Vy^2) / (2 * g), where h is the maximum height reached and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the maximum height will be the same as the height of your mouth. But since the height is not given, we can assume it to be 1 meter for simplicity. So, h = 1 m. Plugging this into the equation, we get:
1 = (Vy^2) / (2 * 9.8).
Next, we need to find the time taken to reach the maximum height, which is half the total time of flight. Let's call this time t1. This can be calculated using the vertical motion equation:
1/2 = Vy * t1 - (1/2) * g * (t1^2).
Now, we can use the time taken to reach the maximum height to calculate the total time of flight (t):
t = 2 * t1.
By rearranging the equation for horizontal distance R and substituting the expressions for Vx and t, we get:
R = V * cos(theta) * t.
Substituting the values we know, we get:
1 = (6 * cos(theta)) * (2 * t1).
Using the trig identity given: 2 * sin(theta) * cos(theta) = sin(2theta), we can rewrite the equation as:
1 = 6 * (2 * sin(theta) * cos(theta)) * sin(2theta).
Now we have an equation to solve for theta. By simplifying it further, we get:
1 = 12 * sin^2(theta) * cos(theta).
To find theta, we would need to solve this equation using algebraic methods or numerical techniques such as graphing or using a calculator or computer software.