A missile is fired upwards from the top of a cliff in a direction making an angle (not known ) with the horizontal. After 20seconds the projectile hits the ground at a point which is 1800metre from the foot of the tower of height 160metres. Find the angle of projection and the initial velocity of the projection and the maximum height?

we know that the path follows

x(t) = v cosθ t
y(t) = 160 + v sinθ t - 4.9t^2

I will assume that the cliff and the tower are the same, otherwise I'm confused.

x(20) = 1800
y(20) = 0

So,

20v cosθ = 1800
20v sinθ = 1800
Clearly θ = π/4 and v = 127.28 m/s

y = 160 + 90t - 4.9t^2
so at t=9.18 the max height of 573.27m is attained.

To solve this problem, we can use the equations of motion for projectile motion. Let's break down the problem into different parts and find the required values.

1. Find the angle of projection:
The angle of projection can be determined using trigonometry. We know that the horizontal distance traveled by the missile is 1800 meters, and the height of the cliff is 160 meters. Let's call the angle of projection θ.

Using trigonometry, we can write:
tan(θ) = (Height of the Cliff) / (Horizontal Distance)
tan(θ) = 160 / 1800

Now, we can find the value of θ by taking the arctan of both sides:
θ = arctan(160 / 1800)
θ ≈ 5.71 degrees

So, the angle of projection is approximately 5.71 degrees.

2. Find the initial velocity of projection:
To find the initial velocity of the projectile, we can use the range equation of projectile motion:
Range (R) = (Initial Velocity)^2 * (sin(2θ)) / g

In this equation, R is the horizontal distance traveled, θ is the angle of projection, and g is the acceleration due to gravity (9.8 m/s^2).

Given the range (R = 1800 m) and the angle of projection (θ ≈ 5.71 degrees), we can rearrange the equation to solve for the initial velocity (u):
u ≈ sqrt((R * g) / sin(2θ))
u ≈ sqrt((1800 * 9.8) / sin(11.42))
u ≈ 208.44 m/s

So, the initial velocity of the projectile is approximately 208.44 m/s.

3. Find the maximum height:
To find the maximum height, we can use the time of flight equation for projectile motion:
Time of Flight (T) = 2 * (Initial Velocity) * sin(θ) / g

Given the time of flight (T = 20 s) and the angle of projection (θ ≈ 5.71 degrees), we can rearrange the equation to solve for the maximum height (H):
H = (Initial Velocity)^2 * (sin(θ))^2 / (2 * g)
H ≈ (208.44)^2 * (sin(11.42))^2 / (2 * 9.8)
H ≈ 2030.81 m

So, the maximum height reached by the missile is approximately 2030.81 meters.

To summarize:
- The angle of projection is approximately 5.71 degrees.
- The initial velocity of the projectile is approximately 208.44 m/s.
- The maximum height reached by the missile is approximately 2030.81 meters.