A projectile is shot from the edge of a cliff h = 125 m above ground level with an initial speed of v0 = 145 m/s at an angle of 37.0° with the horizontal.

(a) Determine the time taken by the projectile to hit point P at ground level.
b) Determine the range X of the projectile as measured from the base of the cliff. (km)
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
m/s (horizontal component)
m/s (vertical component)
(d) What is the the magnitude of the velocity?
(e) What is the angle made by the velocity vector with the horizontal?
° (below the horizontal)

(a) Well, the time taken by the projectile to hit point P can be calculated using the vertical motion equations. Considering the projectile starts at an initial height of 125 m above ground level, we can use the equation for vertical displacement:

h = (v0 * sinθ * t) - (0.5 * g * t^2),

where h is the height, v0 is the initial speed, θ is the angle with the horizontal, t is time, and g is the acceleration due to gravity.

Rearranging this equation, we get:

0 = (0.5 * g * t^2) - (v0 * sinθ * t) + h.

Let's plug in the values:

0 = (0.5 * 9.8 * t^2) - (145 * sin37 * t) + 125.

You might need to solve this equation either using the quadratic formula or using a numerical method. Once you find the positive value of t, that will give you the time taken by the projectile to hit point P.

(b) To determine the range X of the projectile, we can use the horizontal motion equation. This equation relates the distance traveled horizontally to the initial velocity and time:

x = v0 * cosθ * t,

where x is the range and v0, θ, and t are the same as before.

Plugging in the values, we have:

x = 145 * cos37 * t.

Make sure to use the same value of t that you found in part (a) to calculate the range X.

(c) Just before the projectile hits point P, it will have a certain horizontal and vertical velocity. We can find these velocities using the following equations:

Horizontal Velocity (Vx) = v0 * cosθ,
Vertical Velocity (Vy) = v0 * sinθ.

Plug in the values to get the horizontal and vertical components of the velocity.

(d) The magnitude of the velocity (V) can be found using the Pythagorean theorem:

V = √((Vx)^2 + (Vy)^2).

Substitute the values to get the magnitude.

(e) To find the angle made by the velocity vector with the horizontal, you can use the inverse tangent function:

θ' = tan^(-1)(Vy / Vx).

Substitute the values to calculate the angle.

Remember, math may not always be funny, but hey, at least you're getting the answers!

To solve the given problem, we can use the equations of motion for projectile motion. These equations are:

1. Horizontal motion: x = v0x * t
where x is the horizontal displacement, v0x is the initial horizontal component of velocity, and t is the time taken.

2. Vertical motion: y = v0y * t - (1/2) * g * t^2
where y is the vertical displacement, v0y is the initial vertical component of velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.

Now, let's solve the given questions one by one:

(a) Determine the time taken by the projectile to hit point P at ground level.
To find the time taken, we need to find the vertical displacement when the projectile hits the ground (y = 0).
Using equation (2), we can set y = 0 and solve for t:
0 = v0y * t - (1/2) * g * t^2

Simplifying the equation, we get:
(1/2) * g * t^2 = v0y * t

Dividing both sides of the equation by t, we get:
(1/2) * g * t = v0y

Substituting the values, g = 9.8 m/s^2 and v0y = v0 * sin(θ), where θ is the launch angle, we get:
(1/2) * 9.8 * t = 145 * sin(37)

Solving for t, we get:
t = (145 * sin(37)) / (1/2 * 9.8)

Calculating the values, we find:
t ≈ 9.01 s

Therefore, the time taken by the projectile to hit point P is approximately 9.01 seconds.

(b) Determine the range X of the projectile as measured from the base of the cliff.
The range is the horizontal displacement of the projectile when it hits the ground.
Using equation (1), we can substitute the known values and the calculated time (9.01 s):
x = v0x * t
x = v0 * cos(θ) * t
x = 145 * cos(37) * 9.01

Calculating the value, we find:
x ≈ 1147.76 m

To convert it to kilometers, we divide by 1000:
x ≈ 1.15 km

Therefore, the range of the projectile, measured from the base of the cliff, is approximately 1.15 kilometers.

(c) At the instant just before the projectile hits point P, find the horizontal and vertical components of its velocity.
At the instant just before the projectile hits point P, the vertical component of its velocity is 0 m/s, and the horizontal component remains the same as the initial velocity.

The horizontal component of velocity (v0x) can be calculated using the equation:
v0x = v0 * cos(θ)

Substituting the values, we get:
v0x = 145 * cos(37)

Calculating the value, we find:
v0x ≈ 115.97 m/s

Therefore, at the instant just before the projectile hits point P, the horizontal component of its velocity is approximately 115.97 m/s, and the vertical component is 0 m/s.

(d) What is the magnitude of the velocity?
The magnitude of the velocity (v) can be calculated using the Pythagorean theorem:
v = √(v0x^2 + v0y^2)

Substituting the known values, we get:
v = √((145 * cos(37))^2 + (145 * sin(37))^2)

Calculating the value, we find:
v ≈ 145 m/s

Therefore, the magnitude of the velocity is approximately 145 m/s.

(e) What is the angle made by the velocity vector with the horizontal?
The angle (φ) made by the velocity vector with the horizontal can be calculated using the inverse tangent function:
φ = atan(v0y / v0x)

Substituting the known values, we get:
φ = atan((145 * sin(37)) / (145 * cos(37)))

Calculating the value, we find:
φ ≈ 37°

Therefore, the angle made by the velocity vector with the horizontal is approximately 37° below the horizontal.