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How do you find the angle of trajectory to hit a point at a different height with a set velocity?
How do you find the angle of trajectory to hit a point at a different height with a set velocity? You will need to know the horizontal distance to the point!! Think about it. The farther it is to the point, the greater the velocity has to be. The maximum distance for any set velocity is at an angle of 45º. If the velocity is too small, the object will not reach the point no matter even at 45º. So I will use Dh to be the horizontal distance to the point.
I will send you this tonight, but tomorrow I will input values for the velocity, the different height, and the distance horizontal. I will send it again if I find any mistakes.
V = set velocity
Different height =Distance vertical =Dv
Eq. #1 Total distance moved = Vi* t + ½ * at^2
When an object is thrown up into the air, a = acceleration due to gravity =g = -9.8m/s^2
Eq. #2 Distance vertical = V * sin θ* time - (½ * 9.8 * t^2)
½ * 9.8 = 4.9
You can substitute the values of Dv into Eq. #3, to reduce the number of variables. However, I will continue to derive the total equation for the angle.
Eq. #3 Dv = (V * sin θ* t) - (4.9 * t^2)
Range = Distance horizontal = Dh
Eq. #4 Distance horizontal = V * cos θ * time
You can substitute the values of Dh into Eq. #5, to reduce the number of variables. However, I will continue to derive the total equation for the angle.
Eq. #5 Dh = V * cos θ * t
If we solve both equations for V and set them equal to each other, we can solve for t. The time is the same for the Dh and Dv, because these distances are the same for the same one and only object.
From Eq. #3, Dv = (V * sin θ* t) - (4.9 * t^2)
Eq. #6 Dv + (4.9 * t^2) = V * sin θ* t
Eq. #7 V = [ Dv + (4.9 * t^2) ] ÷ (sin θ* t )
From Eq. #5 Dh= V * cos θ * t
Eq. #8 V = Dh ÷ (cos θ * t)
We have 2 equations equal to V. set them equal to each other.
Eq. #9 [ Dv + (4.9 * t^2) ] ÷ (sin θ* t ) = Dh ÷ (cos θ * t)
You may not believe this, but the equation above is a proportion!!
Cross multiply
[ Dv + (4.9 * t^2) ] * (cos θ * t) =Dh * (sin θ * t)
I see factor of t on both sides. Cancel it out.
Eq. #10 [ Dv + (4.9 * t^2) ] * (cos θ) =Dh * (sin θ)
I can multiply (cos θ) * Dv + (4.9 * t^2) ] =
Eq. #11 (Dv * (cos θ) + (cos θ) * 4.9 * t^2) = Dh * (sin θ)
I need t on the left side
Eq. #12 (cos θ) * 4.9 * t^2) = Dh * (sin θ) - (Dv * (cos θ)
Now we will divide both sides by (cos θ * 4.9)
Eq. #13 t^2 = (Dh *sin θ - Dv * cos θ) ÷ (cos θ * 4.9)
You know the value Dh and Dv in Eq. #13.
Solve for t = [(Dh *sin θ - Dv * cos θ) ÷ (cos θ * 4.9)]^0.5
(Square root of t^2)
Eq. #14 t = [(Dh *sin θ - Dv * cos θ) ÷ (cos θ * 4.9)]^0.5
Now substitute the value of Dh, V, and t into the Eq. #5 and find cosθ.
Eq. #5 Dh = V * cos θ * t
Eq. #15 cos θ = Dh ÷ (V * t)
cos θ = Dh ÷ (V * t)
Eq. #16 cos θ = Dh ÷ [ V * [(Dh * sin θ - Dv * cos θ) ÷ (cos θ * 4.9)]^0.5]
Eq. #17 θ = cos^-1 [Dh ÷ [ V * [(Dh * sin θ - Dv * cos θ) ÷ (cos θ * 4.9)]^0.5] ]
The motion of the baseball is parabolic. So, the ball was at the different height twice as it moved along the parabola. The different height would be occurs at 2 places as the baseball follows the parabolic path, once on the way up and once on the way down. The value of the distance horizontal determines the answer; up or down A short Dh would produce a small value of t and a large value of Dh would produce a large value of t.
From Eq. #9 [ Dv + (4.9 * t^2) ] ÷ (sin θ * t ) = Dh ÷ (cos θ * t), we can see that Dh only affects time.
That is why Dh is so critical to the answer for θ.
What else can I help you with?
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