Projectile Motion Problem: Finding Velocity Before Impact

  • Thread starter technology
  • Start date
  • Tags
    Projectiles
In summary, the velocity just before hitting the bottom of the cliff is 71.5 m/s at 61.1° below the horizontal. The answer given on the answer sheet is most likely incorrect.
  • #1
technology
4
0
[SOLVED] Help With Projectiles Problem

Homework Statement


A car rolls down a slope at 30° to the horizontal, and reaches 40 m/s just before plunging off a cliff 180m high. What is its velocity just before hitting the bottom?


Homework Equations


[tex]U_H = vcos\theta[/tex]

[tex]U_V = vsin\theta[/tex]


The Attempt at a Solution


I began the problem by finding intial velocity components:

[tex]U_H = 40cos30^o[/tex]
[tex]U_H = 34.6 m/s [/tex]
[tex]U_V = 40sin30^o[/tex]
[tex]U_V = 20 m/s [/tex]

I then found the time the projectile takes to fall:

[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]-180=-20t-4.9t^2[/tex]
[tex]4.9t^2+20t-180=0[/tex]
[tex]t = 4.35s \mbox{(Using Quadratic Formula)}[/tex]

The horizontal component stays constant at 34.6 m/s

I then calculated the final vertical component

[tex] V_v = U_v + at [/tex]
[tex] V_v = -20 + (-9.8\times4.35) [/tex]
[tex] V_v = 62.6 m/s \mbox {down} [/tex]

I then constructed a vector diagram to calculate the resultant velocity. My working for the resultant velocity was as follows:

[tex] V_R = \sqrt{62.6^2 + 34.6^2} [/tex]
[tex] V_R = 71.5 m/s [/tex]

[tex] tan \theta = \frac{62.6}{34.6} [/tex]
[tex] \theta = 61.1^o [/tex]

Hence, according to my working, the velocity just before hitting the bottom is 71.5 m/s at 61.1° below the horizontal. However, the answer given on the answer sheet to these questions is 65.7 m/s at 72.3° below the horizontal. I am very stuck because I am not sure how to get this answer. Can anyone see where I have gone wrong? Any help would be much appreciated. Thanks very much in advance.
 
Physics news on Phys.org
  • #2
You calculated the time incorrectly using the quadratic formula. However, there is no need to calculate the time of flight, instead, simply use a kinematic equation without time as a variable.
 
  • #3
Ok, I’m still learning this stuff too and apparently I don’t know it as well as I thought. Given this problem I would have solved it the same way, and came up with the same answer as technology.

The only equation I can find that that relates velocity and time (it’s not in the book that I have) is

[tex]

v_y^2 = v_{yo}^2 - 2gy

[/tex]

[tex]

v_y = \sqrt{(-20 m/s)^2 - 2(9.8 m/s^2)(-180)} \approx 62.7~m/s

[/tex]

[tex]

t = \frac{v_{yo} - v_y}{g} = \frac{(-20 m/s) - (62.7 m/s)}{9.8 m/s^2} \approx 4.35 sec

[/tex]
the same as the quadratic formula. =(
 
  • #4
I can see nothing wrong with what either of you did because I am repeatedly getting the same result. Used work-energy and kinematics, both obtained Vf to be 71.6. The answer key is probably incorrect.
 
  • #5
Actually, scratch my previous post. You have calculated the time correctly, it was me that made the mistake (I has a sign wrong :blushing:). Your final answer and solution are correct.
 
Last edited:
  • #6
Actually my solution matches the answer key technology,
heres how:

Vy,f^2 = Vi^2(sin0)-2g(change in y)

Vy,f^2 = 40^2(sin30)- 2(9.83)(180)

Vy,f^2 = 1600(.5)+3538.8

Vy,f^2 = 4338.8

Vy,f = 65.7

Don't second guess yourself hootenanny! :rolleyes:
 
  • #7
robertm said:
Actually my solution matches the answer key technology,
heres how:

Vy,f^2 = Vi^2(sin0)-2g(change in y)

Vy,f^2 = 40^2(sin30)- 2(9.83)(180)

Vy,f^2 = 1600(.5)+3538.8

Vy,f^2 = 4338.8

Vy,f = 65.7

Don't second guess yourself hootenanny! :rolleyes:
Your solution is incomplete, what about the horizontal component of the velocity?
 
  • #8
robertm said:
Actually my solution matches the answer key technology,
heres how:

Vy,f^2 = Vi^2(sin0)-2g(change in y)

Vy,f^2 = 40^2(sin30)- 2(9.83)(180)

you should have

Vy,f^2 = 40^2(sin30)^2 - 2(9.83)(180)

that will get you the 62.7 m/s for Vy,f that RyanSchw and Technology also got.
 
  • #9
Oh crap i see i left off the square on the (sin0)^2. Sorry hootenanny :redface:
 
  • #10
Thanks very much for the help guys, I think the answer key must be wrong. I appreciate the help.
 

1. What is a projectile?

A projectile is any object that is launched into the air and follows a ballistic trajectory (a curved path due to the force of gravity) without any additional propulsion or lift.

2. How do you calculate the trajectory of a projectile?

The trajectory of a projectile can be calculated using the equations of motion, which take into account the initial velocity, angle of launch, and acceleration due to gravity. These equations can be solved using kinematic equations or with the help of computer programs.

3. What factors affect the motion of a projectile?

The motion of a projectile is affected by its initial velocity, angle of launch, air resistance, and the force of gravity. The mass and shape of the projectile can also influence its trajectory.

4. How can you improve the accuracy of a projectile?

To improve the accuracy of a projectile, you can make sure that all the variables, such as initial velocity and angle of launch, are measured and controlled accurately. Also, reducing air resistance by using a streamlined shape or increasing the initial velocity can improve accuracy.

5. What are some real-world applications of projectile motion?

Projectile motion is used in a variety of fields, including physics, engineering, and sports. Examples include calculating the trajectory of a rocket or missile, designing a roller coaster, or predicting the flight of a baseball or golf ball.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
119
  • Introductory Physics Homework Help
Replies
11
Views
871
  • Introductory Physics Homework Help
Replies
11
Views
691
  • Introductory Physics Homework Help
Replies
4
Views
811
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
783
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top