Projectile Motion problem (is my textbook wrong?)

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a stone fired horizontally from a height of 32 meters, landing 135 meters away. Participants are exploring the calculations related to the time of flight and the initial velocity of the stone, questioning the textbook's approach to determining speed.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of time in the air and the subsequent determination of velocity, with some questioning whether the textbook's method calculates average speed. There are alternative approaches suggested for finding initial velocity using different equations. Questions arise about the direction of gravitational acceleration and its effects on horizontal motion.

Discussion Status

Some participants express confusion about the relationship between horizontal and vertical motion, while others clarify that horizontal velocity remains constant in the absence of air resistance. There is acknowledgment of the complexity of projectile motion and the importance of understanding the components of motion.

Contextual Notes

Participants note that they are self-taught beginners in physics and are grappling with the concepts of projectile motion, including the effects of gravity and the interpretation of textbook solutions.

Element1674
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Homework Statement


"A slingshot fires a stone horizontally from a tower 32 metres high, and it lands 135 metres from the foot of the tower. What was the speed at which it left the slingshot?"
vertical Height = 32m
Horizontal Distance = 135m

The Attempt at a Solution


I first calculated the time the rock is in the air using t=sqrt(2s/g). I got 2.55s which is the same as the textbook. However, the textbook then calculates the velocity of the rock using s/t. They get an answer of 53m/s. However this is where I'm confused: isn't this calculating the average speed of the stone?

Alternatively, I calculated the inital velocity of the stone using s=(Vi)(t) + 0.5(a)(t^2). You know all variables in this equation (s=135m, t = 2.55s, a = 9.8m/s^2) so I just rearranged to find the inital velocity.

I'm a self-taught beginner at Physics who just needs a little help with this question :(
 
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Element1674 said:

Homework Statement


"A slingshot fires a stone horizontally from a tower 32 metres high, and it lands 135 metres from the foot of the tower. What was the speed at which it left the slingshot?"
vertical Height = 32m
Horizontal Distance = 135m

The Attempt at a Solution


I first calculated the time the rock is in the air using t=sqrt(2s/g). I got 2.55s which is the same as the textbook. However, the textbook then calculates the velocity of the rock using s/t. They get an answer of 53m/s. However this is where I'm confused: isn't this calculating the average speed of the stone?

Alternatively, I calculated the inital velocity of the stone using s=(Vi)(t) + 0.5(a)(t^2). You know all variables in this equation (s=135m, t = 2.55s, a = 9.8m/s^2) so I just rearranged to find the inital velocity.

I'm a self-taught beginner at Physics who just needs a little help with this question :(

In what direction (horizontal or vertical) does the acceleration due to gravity apply? Does the rock change speed in the horizontal direction?
 
"In what direction (horizontal or vertical) does the acceleration due to gravity apply? Does the rock change speed in the horizontal direction? "

Both? I guess if you're throwing something horizontally wouldn't it slowly fall as a result of the gravitational force? I honestly don't know :( And wouldn't the rock eventually slow down? Or is it because it is constantly going the same horizontal speed and eventually hits the ground because of gravity pushing down on it?

That sounded confusing: I am trying to say that if the rock is constantly going at the same horizontal speed what causes it to fall and stop?
 
Gravity accelerates the stone in the vertical direction but there is nothing that would cause it to accelerate in the horizontal direction. So the horizontal component of the stones velocity is constant (ignoring air resistance).
 
Yes the rock keep on moving in horizontal and vertical motion until it is stopped by the ground.
It's not the average motion but the sum of 2 motions(vectors), vertical and horizontal motion.
 
This makes much more sense now! Thanks! :)
 
There are sereral variants of this question that come up in exams. They typically involve cannon balls being fired up in the air or rolled of the edge of a table of some such. You get to write equations for the horizontal and vertical velocity and are asked to solve them to find some unknown. The first few times you get these problems it may appear as if you don't have enough information to solve the equations but the trick is to look to see what you can get from the start or end conditions. For example if you fire a cannon ball up into the air at an angle there will be a point at which the vertical component of it's velocity falls to zero and it starts to come down again even if the horizontal velocity is a constant 1000mph. Its these extra/end conditions that are usually the key to solving the problem.
 
Another question: "You drive a car 45 miles north 30° west. How much further west are you?"

I drew it so the angle is 30 degrees with the x-axis and then used (cos 30)(45) to find the distance west. My textbook uses cos 60 x 45! :( why am I incorrect?
 

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