Finding angle of projection

In summary, the problem involves a projectile being fired at an angle towards the roof of a tunnel with a given velocity. The goal is to determine the angle of projection, time of flight, and range of the projectile. The correct method involves finding the y-component of the initial velocity and setting up equations of kinematics on the y-axis to solve for the angle and time.
  • #1
curiosity colour
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0

Homework Statement


A projectile is fired with a velocity of 60 m per second at an angle, theta, from a horizontal ground towards the roof of a tunnel of height 5m. If the projectile just barely touches the roof, determine the
a) angle of projection
b) time of the flight
c) range of the projectile

Homework Equations


2-Dimensional Equations [/B]

The Attempt at a Solution

s
I had tried to find it by finding the velocity Y componet first, using v square=u square +2as, after i found it, I use tan theta= y/x, and the answer came out were wrong. And now i have no clue how to do it, please help me[/B]
 
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  • #2
Your method sounds right. Perhaps you could show the detail so that we can check your calculations?

Edit: though now I do it myself, your tan(θ)=y/x looks suspicious.
After you've found Y, the vertical speed, what do you do next?
 
Last edited:
  • #3
The projectile will follow a curved path - not a straight line .

Have a look at the way this problem was solved . You can use similar methods to solve your problem .
 
  • #4
Merlin3189 said:
Your method sounds right. Perhaps you could show the detail so that we can check your calculations?

Edit: though now I do it myself, your tan(θ)=y/x looks suspicious.
After you've found Y, the vertical speed, what do you do next?

yes, you're correct. After you point it out, I look over my drawing, and found out i put the initial velocity at wrong axis(x axis), which should be at the z axis. Now i redraw it, I'm believe this is right. And now i redo it with sin (θ)=y/z, I'm still getting the wrong answer.
https://goo.gl/photos/9CtTGndLvumqJPPEA
Edit: it's seem my picture is not displaying, I'm looking into it.
 
Last edited:
  • #5
Consider the y-axis first

It says that the object barely touches the roof, that means that the maximum height is 5 m
Also at the maximum height the y component of the velocity is 0

Set the two equations of kinematics on the y-axis with those information, you should get a system of two equations with variables the angle theta and the time
 
  • #6
Cozma Alex said:
Consider the y-axis first

It says that the object barely touches the roof, that means that the maximum height is 5 m
Also at the maximum height the y component of the velocity is 0

Set the two equations of kinematics on the y-axis with those information, you should get a system of two equations with variables the angle theta and the time
thx! I got the answer
 

What is the angle of projection?

The angle of projection is the angle at which an object is projected or launched from a specific point. It is measured from the horizontal plane.

Why is it important to find the angle of projection?

Finding the angle of projection is important in determining the trajectory and distance of a projectile. It can also help in optimizing the launch angle for maximum distance or height.

How is the angle of projection calculated?

The angle of projection can be calculated using trigonometric functions such as sine, cosine, and tangent. It also depends on the initial velocity, acceleration due to gravity, and the desired distance or height.

What factors affect the angle of projection?

The angle of projection can be affected by the initial velocity, air resistance, and the launch point's height. The angle also changes depending on the desired distance or height of the projectile.

How can the angle of projection be measured in real-life situations?

In real-life situations, the angle of projection can be measured using tools such as a protractor or a clinometer. It can also be estimated by observing the trajectory of the projectile and adjusting the launch angle accordingly.

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