Projectile motion of 2 dimensions

In summary, the problem asks for the maximum height reached by a rock thrown from a 17.0 m tall building with a velocity of magnitude 24.0 m/s at an angle of 33.0° above the horizontal. The solution involves finding the y-component of the velocity at the maximum height, calculating the time at which the rock hits the ground, and using that time to find the velocity and horizontal distance at the point of impact. The incorrect answers provided in the conversation suggest a misunderstanding of the initial velocity components and the equations used to solve the problem. More effort and understanding are needed to correctly solve the problem.
  • #1
Musicman
21
0
A man stands on the roof of a 17.0 m tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 33.0° above the horizontal. You can ignore air resistance.

(a) Calculate the maximum height above the roof reached by the rock.

(b) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

(c) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.


Ok I started out by doing 24sin(33) and 24cos(33). Then I did 17=0+23.99t+1/2(-9.8)t^2.
I got wrong answers for all 3 parts. I got 23.99 m for part A, for part B I got 44.21 m/s, and part C i got apprx 30 m. Can someone confirm my answers and not just show the answer alone, i need to know HOW to do it.
 
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  • #2
Hey folks,

Just a reminder of the PF Global Guidelines that we all agreed to, under Homework Help:

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

I've added the color for emphasis. Please don't post full solutions to problems. Helping is good. Doing the problem for the questioner is bad.

Thanks,

Tom
 
  • #3
Musicman said:
(a) Calculate the maximum height above the roof reached by the rock.

What is the y-component of the velocity when the rock reaches its maximum height?

(b) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

You have functions that describe both the position and the velocity of the rock as functions of time. Because you know the coordinates of the rock just before it hits the ground, you should be able to find the time at which it hits the ground. Then you should be able to find the velocity at that time.

(c) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

You're being asked about the x-coordinate when the rock hits the ground. You should know what time that is after completing part b.

Ok I started out by doing 24sin(33) and 24cos(33). Then I did 17=0+23.99t+1/2(-9.8)t^2.

Where did the 23.99 come from? You're supposed to have [itex]v_{0y}[/itex] in that spot, but [itex]v_{0y}[/itex] certainly doesn't equal 23.99 m/s.

Try to follow the hints I gave. If you're still stuck then post what you've got and we'll try to get you un-stuck.
 

FAQ: Projectile motion of 2 dimensions

What is projectile motion of 2 dimensions?

Projectile motion of 2 dimensions is a type of motion in which an object is projected into the air and moves along a curved path under the influence of gravity. The object's motion is affected by both horizontal and vertical components.

What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by the initial velocity, angle of projection, and the force of gravity. Other factors such as air resistance and wind can also play a role in the motion.

How is the horizontal and vertical components of a projectile's motion calculated?

The horizontal component of a projectile's motion is calculated using the formula: x = v * t, where v is the initial velocity and t is time. The vertical component can be calculated using the formula: y = v * sin(theta) * t - (1/2) * g * t^2, where theta is the angle of projection and g is the acceleration due to gravity.

What is the maximum height of a projectile?

The maximum height of a projectile can be calculated using the formula: Hmax = (v * sin(theta))^2 / (2 * g), where v is the initial velocity and theta is the angle of projection. This is the highest point on the projectile's trajectory.

How does the range of a projectile change with different angles of projection?

The range of a projectile is the horizontal distance it travels before hitting the ground. It is directly affected by the angle of projection, with a larger angle resulting in a longer range. The optimal angle for maximum range is 45 degrees.

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