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Physics2341313
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I've put the problem statement below and worked it out. I typically don't post questions like this as they're a lot to go through, but I am wondering if I have worked the problem correctly as my book does not have the solution and I feel like I am not understand the material correctly.
1. Homework Statement
A shell is fired upward from the top of a building at an angle of \pi /3 with the horizontal. Its initial velocity is 60 m/s, and assume it is fired from the point (0, 500) when t=0 and x''(t) = 0 and y''(t)=-9.8 m/s^2 while 0 \le t \le t_2 is the time when the shell hits the ground.
Given that x''(t) = 0 and y''(t) = -9.8 we have r''(t) = <0, -9.8> integrating the vector-valued function we have r'(t) = <C, -9.8t + C> \rightarrow r'(t) = <60cos(\pi/3), -9.8t + 60sin(\pi/3)> using the initial velocity given. Integrating again to get the position we have r(t) = <60cos(\pi/3)t + C, -4.9t^2 + 60sin(\pi/3)t + C> r(t) = <60cos(\pi/3)t, -4.9t^2 + 60sin(\pi/3)t + 500> using the point (0, 500) given.
Now, are the parametric equations x(t) and y(t) for r(t)=<x(t),y(t)> just x(t) = 60cos(\pi/3)t and y(t) = -4.9t^2 + 60sin(\pi/3) + 500?
Solving for the time the shell reaches the maximum height will be the time at which y'(t) = 0? so we have: -9.8t + 60sin(\pi/3) = 0 \Rightarrow t = 60sin(\pi/3)/9.8 = 5.3 s
The maximum height will then be y(5.3) = -4.9(5.3)^2 + 60sin(\pi/3)(5.3) + 500 = 637.75 m
The time the shell hits the ground is when y(t) = 0? so we have: 0 = -4.9t^2 + 60sin(\pi/3)t + 500 \Rightarrow t = 10.11
Horizontal distance: x(10.11) = 60cos(\pi/3)(10.11) = 303.3 m
The speed of shell at impact:
||r'(10.11)|| = \sqrt{(60cos(\pi/3))^2 + (-9.8(10.11) + 60sin(\pi/3))^2} = 55.86 m/s
Is this the correct way to calculate the speed? As speed is just the magnitude of the velocity?
1. Homework Statement
A shell is fired upward from the top of a building at an angle of \pi /3 with the horizontal. Its initial velocity is 60 m/s, and assume it is fired from the point (0, 500) when t=0 and x''(t) = 0 and y''(t)=-9.8 m/s^2 while 0 \le t \le t_2 is the time when the shell hits the ground.
The Attempt at a Solution
Given that x''(t) = 0 and y''(t) = -9.8 we have r''(t) = <0, -9.8> integrating the vector-valued function we have r'(t) = <C, -9.8t + C> \rightarrow r'(t) = <60cos(\pi/3), -9.8t + 60sin(\pi/3)> using the initial velocity given. Integrating again to get the position we have r(t) = <60cos(\pi/3)t + C, -4.9t^2 + 60sin(\pi/3)t + C> r(t) = <60cos(\pi/3)t, -4.9t^2 + 60sin(\pi/3)t + 500> using the point (0, 500) given.
Now, are the parametric equations x(t) and y(t) for r(t)=<x(t),y(t)> just x(t) = 60cos(\pi/3)t and y(t) = -4.9t^2 + 60sin(\pi/3) + 500?
Solving for the time the shell reaches the maximum height will be the time at which y'(t) = 0? so we have: -9.8t + 60sin(\pi/3) = 0 \Rightarrow t = 60sin(\pi/3)/9.8 = 5.3 s
The maximum height will then be y(5.3) = -4.9(5.3)^2 + 60sin(\pi/3)(5.3) + 500 = 637.75 m
The time the shell hits the ground is when y(t) = 0? so we have: 0 = -4.9t^2 + 60sin(\pi/3)t + 500 \Rightarrow t = 10.11
Horizontal distance: x(10.11) = 60cos(\pi/3)(10.11) = 303.3 m
The speed of shell at impact:
||r'(10.11)|| = \sqrt{(60cos(\pi/3))^2 + (-9.8(10.11) + 60sin(\pi/3))^2} = 55.86 m/s
Is this the correct way to calculate the speed? As speed is just the magnitude of the velocity?
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