Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Kinematics Projectile Motion Problem
Reply to thread
Message
[QUOTE="joeyoung1996, post: 5001165, member: 541628"] [h2]Homework Statement [/h2] A ball launched from ground level lands 2.85s later on a level field 48 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.) [h2]Homework Equations[/h2] ##x=x_0+v_{0x}t+\frac{1}{2}at^2## ##y=y_0+v_{0y}t+\frac{1}{2}at^2## [h2]The Attempt at a Solution[/h2] So, I worked out this problem seemingly without an issue, but WebAssign keeps telling me the answers are wrong, so maybe a bit of new eyes can point out what I've done wrong. I start by attempting to solve for the x component of the velocity vector: [CENTER]##x=x_0+v_{0x}t+\frac{1}{2}at^2## ##48m=0m+v_{0x}(2.85s)+\frac{1}{2}(0)(2.85s)^2## ##48m=v_{0x}(2.85s)## ##v_{0x}=\frac{48m}{2.85s}## ##v_{0x}=16.84\frac{m}{s}## [/CENTER] I then found the y component of the velocity: [CENTER]##y=y_0+v_{0y}t+\frac{1}{2}at^2## ##0m=0m+v_{0y}(2.85s)+\frac{1}{2}(-9.8\frac{m}{s^2})(2.85s)^2## ##0=v_{0y}(2.85s)-13.96m## ##v_{0y}=\frac{13.96m}{2.85s}## ##v_{0y}=4.9\frac{m}{s}## [/CENTER] Now that I have both components, I use the Pythagorean Theorem to find the magnitude of the velocity: [CENTER]##||\vec{v_0}|| = \sqrt{v_{0x}^2+v_{0y}^2}## ##||\vec{v_0}||= \sqrt{(16.84\frac{m}{s})^2+(4.9\frac{m}{s})^2}## ##||\vec{v_0}||=17.5\frac{m}{s}## [/CENTER] Finally, I use these to find the angle at which the ball was fired: [CENTER] ##sinx=\frac{4.9}{17.5}## ##x=sin^{-1}(\frac{4.9\frac{m}{s}}{17.5\frac{m}{s}})## ##x=16.22^{\circ}## [/CENTER] I figured I have just made a simple, stupid mistake, but for the life of me, I cannot find it. :oldgrumpy: :headbang: [CENTER][/CENTER] [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Kinematics Projectile Motion Problem
Back
Top