What Is the Optimal Angle to Hit a 5m High Target with a 60 m/s Projectile?

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Homework Help Overview

The problem involves determining the optimal angle for a projectile launched at 60 m/s to just touch the roof of a tunnel that is 5 m high. The discussion centers around projectile motion and the application of kinematic equations in two dimensions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial attempts to find the angle of projection using vertical and horizontal components of velocity. There are questions about the correctness of the approach, particularly regarding the use of trigonometric relationships and the setup of the coordinate axes.

Discussion Status

Some participants have provided guidance on focusing on the vertical motion and the conditions at maximum height. There is an acknowledgment of the need to set up equations based on kinematics, but no consensus has been reached on the specific method to solve the problem.

Contextual Notes

Participants note that the projectile must reach a maximum height of exactly 5 m, and there is a mention of the vertical component of velocity being zero at this height. The original poster expresses confusion over their calculations and the setup of the problem.

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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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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:
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 .
 
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:
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
 
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
 

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