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:
- 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.
- 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.
- 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:
- 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).
- Free fall acceleration is constant over the range of motion and is directed downward
- Unless otherwise stated, the affect of air resistance is negligible
Position vector of a projectile is r-f = ri + Vit + 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:
- The origin is the ground level from which the projectile is launched from and the positive Y direction is straight upwards.
- Vo= 65 m/s
- Theta= 37 degrees
- acceleration of the Y component = -g
- Y0= 125 meters (displacement)
- Vyo= Vosinφ
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=Yo+Vyot+1/2ayt2. plugging in the knowns from the first part into this equation and rearranging this equation for time (t) we get:
t=-vosinφ±√(vo2sin2φ-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=Vxt = V0cosφt = (65 m/s)(cos37)(10.4s) = 541 m
C) The horizontal component and vertical component velocity:
Vx= V0cosφ= morning (65m/s)(cos37)= 51.9 m/s
Vy=Vyo+at= Vosinφ-gt= (65m/s)(sin37)-(9.8m/s2)(10.4s) = -63.1 m/s
D) The magnitude of the velocity vector
v=√(vx2)+(vy2) = √(51.9m/s)2+(-63.1m/s)2= 81.7 m/s
E) The direction of the velocity vector:
φ=tan-1(Vy)/(Vx) = 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.
Yf2= Vyi2+ 2aΔy ----> (rearrange for Y max). Ymax= Vo2sin2φ/2g = (65 m/s)2sin2(37)/2(9.8m/s2)= 78.1m