Projectile Motion - object that is shot at an angle of 55 degrees

• Yellowkies_3275
In summary: You are a life, saver thank you very much for your explanationsIn summary, the conversation discusses a projectile motion problem where an object is shot at an angle of 55 degrees with a velocity of 35 m/s. The object is launched from the ground and is meant to hit a button elevated by 32 feet. The equations for Delta x and delta y are provided, but the solution is not clear. Through further discussion and guidance, the problem is solved by separating the initial velocity into horizontal and vertical components and adjusting the equation for delta y to account for acceleration due to gravity. The -9.8 m/sec^2 value is used to prevent the object from accelerating into outer space. The problem is successfully solved with the given information.
Yellowkies_3275

Homework Statement

Projectile motion
I have an an object that is shot at an angle of 55 degree's. (the velocity listed next to it was 35 m/s, which was not listed as Vo but I think I have to assume it is?) It is launched from the ground. it is meant to land to hit a button that is elevated by 32 feet. we have to find the distance away from the button to launch the object in order to hit the button.

Homework Equations

Delta x= Vo * t
delta y = 1/2gt^2 + Vo * t

The Attempt at a Solution

I tried a bunch of things that did not help me come to any sort of conclusion about the distance (x)

Last edited:
Please show us what you've tried.

For the ## x=v_x t ##, you need to have ## v_x=v_o \cos(55^{\circ}) ##. Do you see why? ## \\ ## For the y equation, you need ## y=-\frac{gt^2}{2}+v_{yi} t ## where ## v_{yi}=v_o \sin(55^{\circ}) ##. The acceleration due to gravity is downward=thereby the minus sign. ## \\ ## Meanwhile, an important part you are missing is to separate the initial velocity ## v_o ## at an angle of ## 55^{\circ} ## into horizontal and vertical components. I have shown you this result in the equations above. ## \\ ## Now that I got you started, can you figure out what to do to answer the question they are asking? Here's a hint: Can you solve for ## y ## as a function of ## x ##? ## \\ ## Presently, the equations you have go through the point ## x=0,y=0 ## at ## t=0 ##. What is ## x ## when ## y=32 ## ? You need to determine the equation that gives you ## y ## as a function of ## x ##, and then you can answer that question.

Last edited:
Yellowkies_3275
gneill said:
Please show us what you've tried.
sorry i didn't fill it out properly. i think i have a solution to the question now though, so i was wondeering if this thread could be deleted

Yellowkies_3275 said:
sorry i didn't fill it out properly. i think i have a solution to the question now though, so i was wondeering if this thread could be deleted
After a thread has received replies, particularly where contributors have put in significant effort, our policy is to not remove the thread so that other members can benefit from the information.

Charles Link
Yellowkies_3275 said:

Homework Statement

Projectile motion
I have an an object that is shot at an angle of 55 degree's. (the velocity listed next to it was 35 m/s, which was not listed as Vo but I think I have to assume it is?) It is launched from the ground. it is meant to land to hit a button that is elevated by 32 feet. we have to find the distance away from the button to launch the object in order to hit the button.

Homework Equations

Delta x= Vo * t
delta y = 1/2gt^2 + Vo * t

The Attempt at a Solution

I tried a bunch of things that did not help me come to any sort of conclusion about the distance (x)

Be very careful. If ##y## measures height above the ground, your formula ##\Delta y = v_0 t + \frac{1}{2} g t^2## will have the object accelerating rapidly upward, until it goes into outer space (because, conventionally, ##g## is a positive constant).

Ray Vickson said:
Be very careful. If ##y## measures height above the ground, your formula ##\Delta y = v_0 t + \frac{1}{2} g t^2## will have the object accelerating rapidly upward, until it goes into outer space (because, conventionally, ##g## is a positive constant).
How do I change the equation to keep it from accelerating into outer space?

Yellowkies_3275 said:
How do I change the equation to keep it from accelerating into outer space?
You put a minus sign in front of the ## \frac{g \, t^2}{2} ## term. ## g=+9.8 ## m/sec^2, but ## a=-9.8 ## m/sec^2. ## \\ ## Otherwise, your equation for ## y ## has the form ## y=At^2+Bt ## with ## A>0 ## , so that ## y ## will get increasingly larger with time ## t ##.

Yellowkies_3275
Charles Link said:
You put a minus sign in front of the ## \frac{g \, t^2}{2} ## term. ## g=+9.8 ## m/sec^2, but ## a=-9.8 ## m/sec^2. ## \\ ## Otherwise, your equation for ## y ## has the form ## y=At^2+Bt ## with ## A>0 ## , so that ## y ## will get increasingly larger with time ## t ##.
You are a life, saver thank you very much for your explanations

Charles Link

What is projectile motion?

Projectile motion is the motion of an object that is projected into the air at an angle and then moves along a curved path under the influence of gravity.

What factors affect projectile motion?

The factors that affect projectile motion include initial velocity, angle of projection, air resistance, and gravitational force.

How do you calculate the range of a projectile?

The range of a projectile can be calculated using the formula R = (v^2 * sin(2θ))/g, where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

What is the maximum height reached by a projectile?

The maximum height reached by a projectile can be calculated using the formula H = (v^2 * sin^2(θ))/2g, where H is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

How does air resistance affect projectile motion?

Air resistance can decrease the range and height of a projectile by slowing it down as it travels through the air. It also causes the projectile to follow a slightly curved path rather than a perfectly parabolic one.

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