a particle is projected at an angle tita to the horizontal with an initial velocity U.if the maximum horizontal distance travelled is 20m and the greatest height reached is 10m.find U and Tita

the horizontal distance is

x(t) = u cosθ t
The time taken to go 20m is thus
t = 20/(u cosθ)

The trajectory is
y(t) = u sinθ t - g/2 t^2
We know that y=0 at 0 and 20/(u cosθ), so the max height is reached at 10/(u cosθ). So,

u sinθ * 10/(u cosθ) - g/2 (100/(u^2 cos^2 θ) = 10

y = tanθ x - g/(2 (u cosθ)^2) x^2

We know that y(20) = 0 and y(10) = 10. So, approximating g as 5 m/s^2,

20tanθ - 500/(u cosθ)^2 = 0
10tanθ - 125/(u cosθ)^2 = 10

u = 5/2 √10
tanθ = (√5-1)/2

Initial vertical speed Vi = U sin theta

constant horizontal speed = u = U cos theta

horizontal problem
20 = U cos theta * T where T is time in air

vertical problem
v = Vi - g t
v = 0 at the top
so
t at top = Vi/g
g = 9.8 m/s^2 on earth
so
t at top = Vi/9.8

now at the top at 10 meters
h = 0 + Vi t -4.9 t^2
10 = Vi t - 4.9 t^2
10 = Vi^2/9.8 - 4.9 Vi^2/96
10 = Vi^2 (.051)
Vi = 14 m/s
t at top = 14/9.8 = 1.43 s
total time in air = 2t = 4.29 s
20 = u (4.29)
so u = 4.67 m/s
so
U = sqrt (4.67^2+14^2) = 14.8 m/s
and
tan theta = 14/4.67
theta = 71.6 degrees

t at top = 14/9.8 = 1.43 s

total time in air = 2t = 2.86 s
20 = u (2.86)
so u = 6.99 m/s
so
U = sqrt (6.99^2+14^2) = 15.64 m/s
and
tan theta = 14/6.99
theta = 63.5 degrees

To find the initial velocity (U) and the launch angle (θ) of the particle, we first need to understand the motion of the particle.

Let's break down the motion into horizontal and vertical components. The horizontal component is constant and unaffected by gravity, while the vertical component is affected by gravity.

Horizontal component:
The maximum horizontal distance traveled is equal to the range (R). We are given that R = 20m.

Vertical component:
The greatest height reached by the particle is equal to the maximum vertical displacement (H). We are given that H = 10m.

Now, let's derive the equations for the horizontal and vertical components of motion.

Horizontal component:
The horizontal distance traveled is given by the formula:
R = U * cos(θ) * t
- Where U is the initial velocity, θ is the angle of projection, and t is the time of flight.

Vertical component:
The maximum height reached by the particle is given by the formula:
H = (U * sin(θ))^2 / (2 * g)
- Where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now that we have the formulas, we can use them to find U and θ.

First, let's find U:

1. We know that R = 20m. From the horizontal component equation, we have:
20 = U * cos(θ) * t

Since we only have one equation with two unknowns (U and θ), we need another equation. Let's solve for t:

2. We can find t using the formula for the time of flight:
t = (2U * sin(θ)) / g

Now, let's substitute this expression for t into equation 1:

20 = U * cos(θ) * [(2U * sin(θ)) / g]

Simplifying, we get:
20 = 2 * U^2 * (sin(θ) * cos(θ)) / g

Using the trigonometric identity sin(2θ) = 2sin(θ) * cos(θ), we can rewrite the equation:

20 = (U^2 * sin(2θ)) / g

This equation gives us a relationship between U and θ.

Now, let's find θ:

3. We know that H = 10m. From the vertical component equation, we have:
10 = (U * sin(θ))^2 / (2 * g)

Rearranging this equation, we get:
20 * g = (U^2 * sin(θ))^2

Taking the square root of both sides to eliminate the square, we get:
√(20 * g) = U * sin(θ)

Squaring both sides again, we get:
20 * g = U^2 * sin^2(θ)

Now, we can substitute this expression for U^2 * sin^2(θ) into equation 2:

20 = (U^2 * sin(2θ)) / g

Substituting the value of U^2 * sin^2(θ) from equation 3, we have:
20 = (20 * g * sin(2θ)) / g

Simplifying, the g on both sides cancels out:
20 = 20 * sin(2θ)

Finally, dividing both sides by 20, we get:
1 = sin(2θ)

Now, to find the angle θ, we need to find the inverse sine (sin^-1) of 1/2:

2θ = sin^-1(1)
θ = sin^-1(1) / 2

Using a calculator, we find that θ ≈ 30°.

Now that we have θ, we can substitute it into equation 3 to solve for U:

U * sin(θ) = √(20 * g)
U = √(20 * g) / sin(θ)

Substituting the values of g (9.8 m/s^2) and θ (30°) into the equation, we can calculate U:

U = √(20 * 9.8) / sin(30°)
U = √(196) / 0.5
U = 14 m/s

Therefore, the initial velocity (U) is 14 m/s and the launch angle (θ) is 30°.