a particle is projected at angle tita to the horizontal with initial velocity u.if the horizontal distance and maximum height reached are 20m and 10m respectively. find the value of tita and u
H=u^2÷g
Y^2 = Yo^2 + 2g*h = 0
Yo^2 + (-19.6)10 = 0
Yo = 14 m/s = vertical component of initial velocity.
Y = Yo + g*Tr = 0
14 + (-9.8)Tr = 0
Tr = 1.43 s. = Rise time.
Tf = Tr = 1.43 s. = Fall time.
T = Tr+Tf = 1.43+1.43 = 2.86 s. = Time in air.
d = Xo*T = 20
2.86Xo = 20
Xo = 7.0 m/s = horizontal component of initial velocity.
a. Vo = Xo + Yoi = 7 + 14i = 15.7 m/s[63.4o] = Initial velocity.
b. Theta = 63.4o.
To find the value of θ and u, we can use the equations of motion for projectile motion. Projectile motion involves the vertical and horizontal components of motion separately.
Let's derive the equations we need step by step:
1. Vertical motion:
At the highest point, the vertical component of velocity becomes 0. We can use the equation:
v = u - gt,
where v is the final vertical velocity (0 in this case), u is the initial vertical velocity (v₀sinθ), g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight.
Plugging in the values, the equation becomes:
0 = u - 9.8 × t
2. Horizontal motion:
The horizontal distance traveled (20 m) is given by:
x = u * t * cosθ,
where x is the horizontal distance, u is the initial velocity, t is the time of flight, and θ is the angle of projection.
Rearranging the equation, we get:
t = x / (u * cosθ)
3. Maximum height:
The maximum height (h) reached is given by:
h = (v₀sinθ)² / (2g),
where h is the maximum height, v₀ is the initial vertical velocity, θ is the angle of projection, and g is the acceleration due to gravity.
Plugging in the values, the equation becomes:
10 = (u * sinθ)² / (2 * 9.8)
Simplifying further, we get:
10 = (u² * sin²θ) / 19.6
Using these equations, we can solve for θ and u:
From equation 1:
0 = u - 9.8t ---(1)
From equation 2:
t = 20 / (u * cosθ) ---(2)
From equation 3:
10 = (u² * sin²θ) / 19.6 ---(3)
Now, we can substitute equation 1 and equation 2 into equation 3:
10 = ((9.8t)² * sin²θ) / 19.6 [Substituting u = 9.8t from equation 1]
10 = t² * sin²θ
Next, we can simplify equation 3:
10 = t² * (1 - cos²θ) [Using the trigonometric identity sin²θ + cos²θ = 1]
10 = t² - t² * cos²θ
10 = t²(1 - cos²θ)
Now, we can substitute equation 2 into the simplified equation 3:
10 = (20 / (u * cosθ))²(1 - cos²θ)
10 = (400 / (u² * cos²θ)) * (1 - cos²θ)
10 = 400 / (u² * cos²θ) - 400 / u²
Multiply the equation by (u² * cos²θ) to eliminate the denominator:
10u² * cos²θ = 400 - 400 * cos²θ
10u² * cos²θ + 400 * cos²θ = 400
Now, rearrange the equation:
10u² * cos²θ + 400 * cos²θ - 400 = 0
This is a quadratic equation in terms of cos²θ. We can solve this equation to find the value of cos²θ. Once we find cos²θ, we can find cosθ using the identity cos²θ + sin²θ = 1 and then calculate θ.
Once we have the value of cosθ, we substitute it back into equation 1 to solve for u.
I apologize for the complexity of this problem, but it involves solving a quadratic equation. If you have a specific value for x or y that you already know, I can help you solve it further.