Calculating Projectile Motion: Rock Kicked From a Building

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Homework Help Overview

The problem involves calculating the distance a rock travels horizontally after being kicked off a building at a specific velocity. The scenario is set in the context of projectile motion, specifically focusing on the horizontal and vertical components of motion under the influence of gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial conditions of the problem, questioning the relationship between horizontal and vertical motion. There are attempts to apply kinematic equations, with some participants expressing uncertainty about how to approach the problem and whether to consider angles in their calculations.

Discussion Status

Some participants have provided guidance on separating the horizontal and vertical components of motion. There is ongoing exploration of how to correctly apply the equations of motion, with some participants sharing their attempts and results, while others are still seeking clarity on the concepts involved.

Contextual Notes

There is mention of confusion regarding the signs in the equations and the treatment of initial velocities, particularly noting that the initial velocity in the vertical direction is zero since the rock is kicked horizontally. Participants are also addressing the need to treat horizontal and vertical motions independently.

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kicking rock of a building...

Homework Statement



A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

Homework Equations



sorry, i don't know how to type some symbols

v^2 = v(initial)^2 + 2a(change in x)
a(of y)= -9.8
v^2(for y) = v(initial of y)^2 + 2a(change in y)


The Attempt at a Solution



not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

The answer is 71.6 meters, for the record.
 
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davidatwayne said:

Homework Statement



A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

Homework Equations



sorry, i don't know how to type some symbols

v^2 = v(initial)^2 + 2a(change in x)
a(of y)= -9.8
v^2(for y) = v(initial of y)^2 + 2a(change in y)


The Attempt at a Solution



not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

The answer is 71.6 meters, for the record.
19.8m/s is the initial velocity and is given in the horizonatl direction. Ther is no y component of the initial velocity. You've got to use your motion equations in each direction, and note thatt in the x direction, there is no acceleration, since no forces act in that direction during the flight.
 
davidatwayne said:

Homework Statement



A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

Homework Equations



sorry, i don't know how to type some symbols

v^2 = v(initial)^2 + 2a(change in x)
a(of y)= -9.8
v^2(for y) = v(initial of y)^2 + 2a(change in y)


The Attempt at a Solution



not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

The answer is 71.6 meters, for the record.


1) initial position and initial velocity
This is a 2-D problem so you have x (horizontal) and y (vertical) components

initial position : x_i = 0 e_x + 64 e_y
initial velocity : v_i = 19.8 e_x + 0 e_y

Gravity works along the y-axis, which you implemented correctly (-9.81)

2) So applying Newton's second law gives us :

a_i = 0 t^2/2 + 19.8 t
a_y = -9.81 t^2/2 + 0 t + 64

Do you understand these equations ?

If so, solve them for t and you will have your answer

marlon
 
I am still not getting it...

I tried using the quadratic formula to solve for the time, using

64 (y) = [19.8 (initial velocity) x t] - (1/2)(9.8)t^2

I came up with an answer of 2.02 seconds.

then I used:

x = [v (initial velocity) x time] + (1/2) x (-9.8) x (time squared)

came up with - 19.98.
which is not the correct answer, nor can a change in x be negative
 
davidatwayne said:
I tried using the quadratic formula to solve for the time, using

64 (y) = [19.8 (initial velocity) x t] - (1/2)(9.8)t^2
You must take care to treat horizontal and vertical motion separately. What's the y-component of the initial velocity? (Recall that the rock is kicked horizontally.)

Also, take care with the signs. Use + for up and - for down. The full equation for the vertical position is:
y_f = y_i + v_{i,y} t - (1/2) (9.8) t^2
 
ok, i think i got it... i was looking for how v for the x direction and v for the y direction were related... thanks for your help guys : )
 

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