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

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The discussion focuses on determining the optimal angle for a projectile launched at 60 m/s to just touch a 5m high target. Participants emphasize the importance of correctly identifying the vertical and horizontal components of the projectile's motion. The maximum height of the projectile must equal 5m, where the vertical component of velocity becomes zero. By setting up kinematic equations for the y-axis, users can derive a system of equations involving the angle of projection and time. Ultimately, one participant successfully solved the problem after correcting their approach.
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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?
 
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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.
 
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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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