I really need help with this physics problem

Question

So the problem is "A projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 65.0m/s at an angle of 37.0 degrees 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. (c)At the instant just before the projectile hits point P, find the horizontal and vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile.

Here is the information I know:

X₀=0m (initial distance of x)

V_x0=65cos(37) (initial velocity of x)

Y₀=125m (initial distance of y)

Y=0m (final distance of y)

V_y0=65sin(37) (initial velocity of y)

g=-9.8 m/s^2 (gravity)

Answer 1

This Problem is a projectile is shot from a height of 125 meters above the ground.

to go to ground level

it is shot at 65.0m/s at an angle of 37.0 degrees so it its initial y velocity component is 65 * sine 37 =39.1 m/s

so the time to get to the level of the it has to go up and down so time up + tie to fall to cliff level time to stop rising 0 = 37-gt or 3.77 sec to time to come down is the same 3.77

the time to get back to cliff level is 2 * 3.77 or 7.4 sec the down ward velocity as it passes the edge of cliff is 39.1 m/s. Maximum height is 4.8*3.77² = 69.6 meters

the equation for distance is D=0.5at² + st in our case D=125 meters

so equation is 125 = 0.5*9.8*t² *s*t this is a quadratic equations and needs to be put in proper form

+

0 = 4.9*t^{2 +}39.1*t + -125l solve for t i get 2.45 seconds to hit ground below cliff

in the down range velocity is cos(37)*65 = 52 m/s

The total time from when the projectile is fired is 4.7 + 2.45 =7.15 sec so the distance from the shooter

7.15*52m/ = 371 meters from the shooter from the base cliff to the shooter is unknown if the shooter is at the edge ( I do not recommend this position to do anything) is shown above.

please comment

Answer 2

So when answering this question you first must look at what form of physics model we will be using based on the wording of our question; in this case since we are dealing with "projectile motion" we must remember that this form of a particle model is with motion in two dimensions and we will potentially be dealing with the the "BIG 5" kinematic equations.

Now with this, even If they apply to this question are not it is important to remember some key facts about two-dimension motion:

1. In two dimensions a particle position is now the position vector (r^-) from the origin of a coordinate system to the particle's location in the XY plane.
2. When a particle accelerates in two dimensions the following is possible: A) the magnitude of the velocity (speed) vector can change with time and (one-dimension property); B) the velocity vector direction can change if the speed magnitude is constant along a curved path; C) the magnitude and direction of the velocity vector can simultaneously change.
3. if the particle only has an X-component of acceleration then the X-direction is modeled as a particle under constant acceleration and the Y-direction is modeled as a particle under constant velocity.

Now with that in mind, we must remember the following for projectile motion:

1. In projectile motion we have two analysis models: one being constant acceleration in the vertical direction and the other model being constant velocity in the horizontal direction with both being related in respect to time (t).
2. Free fall acceleration is constant over the range of motion and is directed downward
3. Unless otherwise stated, the affect of air resistance is negligible

Position vector of a projectile is r^-_f = r_i + V_it + 1/2g^-t² with initial velocity being regards to displacement if no acceleration is present and "g" in regards to acceleration due to gravity. With the initial X and Y components being Vxi=Vjcosθ and Vyi=Vjsinθ.

Now moving to solve this problem we must write down all the known variables:

1. The origin is the ground level from which the projectile is launched from and the positive Y direction is straight upwards.
2. V_o= 65 m/s
3. Theta= 37 degrees
4. acceleration of the Y component = -g
5. Y₀= 125 meters (displacement)
6. V_yo= V_osinφ

A) The time needed to reach the ground: to (this means that the final height will be zero)

Since we are going to be using are kinematic equations we will be using the equation for a particle under constant acceleration with known initial velocity and initial position and Ac in the Y-component direction.

Y=Y_o+V_yot+1/2a_yt^2. plugging in the knowns from the first part into this equation and rearranging this equation for time (t) we get:

t=-v_osinφ±√(v_o²sin²φ-4(-1/2g)(125))/2(-1/2g) = -39.1±63.1/-9.8 = 10.4s and/or -2.45s <---since we know the time component is always a positive quantity we know that the positive value is correct and we ignore the negative value.

B) The horizontal range can be found from the horizontal motion at constant velocity (the is from the top part that we pulled from information we should remember because we are working in the horizontal direction).

Δx=V_xt = V₀cosφt = (65 m/s)(cos37)(10.4s) = 541 m

C) The horizontal component and vertical component velocity:

V_x= V₀cosφ= morning (65m/s)(cos37)= 51.9 m/s

V_y=V_yo+at= V_osinφ-gt= (65m/s)(sin37)-(9.8m/s²)(10.4s) = -63.1 m/s

D) The magnitude of the velocity vector

v=√(v_x²)+(v_y²) = √(51.9m/s)²+(-63.1m/s)²= 81.7 m/s

E) The direction of the velocity vector:

φ=tan^-1(V_y)/(V_x) = tan^-1(-63.1)/(51.9)= -50.6 degrees ( this means that object is moving below the horizontal at 50.6 degrees.

F) Max height of the projectile (this will occur when the Y-component velocity is equaled to zero)

Using the kinematic equation for velocity as a function of position under constant acceleration in which we are not concerned about time.

Y_f²= V_yi²+ 2aΔy ----> (rearrange for Y max). Y_max= V_o²sin²φ/2g = (65 m/s)²sin²(37)/2(9.8m/s²)= 78.1m