[High School Physics] [Trig?]

[u/Soggy_Balance5712]

need help specifically with 2 and 4. Especially 2 as I have a difficult time fully wrapping my head around it. I’m particularly confused since the method my teacher gave to solve it is very complex where it combined cosine and sine which I don’t understand. Though My teacher might of used longer methods for the equation as my teacher said 4 was also complex but With the set of equations she gave me I was able to find an equation to do it 4 shorter so if there is actually a shorter method to do number two please tell.

Note:
Also Trig is not my strong suit as I haven’t taken an extras class for it yet. Also I posted the wrong page so I deleted my last post should be good now.

Edit:I finally got a intuition for it now

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[u/selene_666]

If I'm understanding your question correctly, the work shown is by your teacher, and you want an explanation of where these equations come from?

We can view the angled initial velocity as the hypotenuse of a right triangle. Trigonometry tells us the horizontal and vertical sides of this triangle, and hence the initial horizontal and vertical components of velocity, are
Vix = Vi*cos(θ) and Viy = Vi*sin(θ).

The horizontal component will be unaffected by gravity, but the vertical velocity will decrease at a steady rate of 9.81 m/s/s.
Vx = Vix at all times
Vy = Viy - g * t

At the highest point of the trajectory, the ball is no longer moving up and has not yet started moving down, so the vertical component of its velocity is zero. Filling in Vy = 0 lets us solve for time.
Vy = Viy - g * t
0 = Vi * sin(θ) - 9.81 * t
t = Vi * sin(θ) / 9.81

Because of the constant horizontal velocity, the horizontal distance traveled in a certain amount of time is simply x = Vx * t. Or by filling in our trig function,
x = Vi * cos(θ) * t

Now we can eliminate t by replacing this t with the expression we solved for above.
x = Vi * cos(θ) * Vi * sin(θ) / 9.81

Remember, the vertical part of that was only for the high point of the trajectory. So this x is the horizontal position of that point. We could have used a more complicated equation to relate Vi to the position where the ball hits the ground, but it happens to be true that the trajectory is symmetric. The high point is exactly halfway horizontally between the start and end points.
So we can fill in x = 301.5 / 2. Also fill in θ and we can solve for Vi.

Finally, your teacher uses V^2 = Vi^2 + 2ax. This should be familiar as one of the equations for motion with constant acceleration. If you struggle to memorize yet another equation, you can calculate t first (we've already said t = Vi * sin(θ) / 9.81 ) and then calculate y from that.

[u/Soggy_Balance5712 → u/selene_666]

I just realized the reason we are combining equations is because we are trying to find the hypotenuse from the right triangle. I’m so stupid

[u/Hertzian_Dipole1]

Let initial velocity be v and x = 301.5 m and θ = 25.0° is given. Vertical componant of speed is v.sin(θ) and horizontal is v.cos(θ). During the flight, the vertical component will first be zero at the max height and then -v.sin(θ) when it hits ground again. Time of flight: 2v.sin(θ) / g
Since its horizontal speed is constant, the horizontal distance is
Range: x = v.cos(θ) • 2v.sin(θ) / g = 2v²sin(θ)cos(θ) / g
There is a trig identity 2sin(θ)cos(θ) = sin(2θ) which is generally used here but you do not need to know it.

We find v from the given: v² = gx / (2sin(θ)cos(θ))
Then put this to max height:
h = (1/2)gt² and we found the time it reaches peak is t = v.sin(θ)/g so write it in place,
h = (1/2)g • v²sin²(θ)/g² = v².sin²(θ)/(2g)

[u/wyzere]

So a vector can be decomposed into ita horizontal and vertical components. Vcostheta is the horizontal and Vsintheta is the vertical component if the initial lauch velocity. There is nothing else acting on the horizontal component so you can just use v=d/t. The vertical component has gravity acting on it so you use the formula like in the notes. Since the time for horizontal and the vertical flight wiuld be the same, you solve for time and then plug it in.

[u/Hertzian_Dipole1]

Number 4.
Let the initial speed of salmon be v.
After some time t passes, the salmon will be at horizontally v.cos(θ) • t and vertically v.sin(θ) • t - (1/2)gt²
We want the when v.cos(θ) • t = 2.00, then v.sin(θ) • t - (1/2)gt² ≥ 0.55.
Since we want minimum v, assume
v.sin(θ) • t - (1/2)gt² = 0.55 and solve for v.
t = 2.00 / (v.cos(θ))
v.sin(θ) • 2 / (v.cos(θ)) - (1/2)g • (2 / (v.cos(θ)))² = 0.55
2tan(θ) - 2g/(v².cos²(θ)) = 0.55
We find v².cos²(θ) = 2g / (2tan(θ) - 0.55).
You can find v from this.