Salazar Posted September 19, 2012 Share Posted September 19, 2012 Help, I am having a difficultly setting up this problem. A baseball is hit at a height H=1.00m and then caught at the same height. It travels alongside a wall, moving up past the top of a wall 1.00s after it is hit and then down past the top of the the wall 4 seconds later, at a distance D=50.0m farther along the wall. Find : (A) Horizontal distance traveled from hit to catch. Link to comment Share on other sites More sharing options...
swansont Posted September 19, 2012 Share Posted September 19, 2012 What do you have so far for a solution? Link to comment Share on other sites More sharing options...
Salazar Posted September 19, 2012 Author Share Posted September 19, 2012 I only have the initial velocity (Vi2) after the time interval of 3 seconds with a traveled distance of 50.0m (Vi2)(cosα)=50m/3s= 16/67m/s). This helped me get the initial velocity (Vi) after the ball was hit (Vo=Vf-at; Vo=16.67m/s+(9.8m/s^2)(1s)=26.47m/s). But I have no angle so how do I find the x-component of velocity to get the distance traveled in that one second interval (with the equation (Vx=Vo(cosαα)? Link to comment Share on other sites More sharing options...
ACUV Posted September 19, 2012 Share Posted September 19, 2012 (edited) Salazar, There is no resistance to horizontal motion. The motion will be constant. The time taken will be proportional to the distance travelled. Edited September 19, 2012 by ACUV Link to comment Share on other sites More sharing options...
swansont Posted September 20, 2012 Share Posted September 20, 2012 I only have the initial velocity (Vi2) after the time interval of 3 seconds with a traveled distance of 50.0m (Vi2)(cosα)=50m/3s= 16/67m/s). This helped me get the initial velocity (Vi) after the ball was hit (Vo=Vf-at; Vo=16.67m/s+(9.8m/s^2)(1s)=26.47m/s). But I have no angle so how do I find the x-component of velocity to get the distance traveled in that one second interval (with the equation (Vx=Vo(cosαα)? The distance along the wall is the x motion, so it travels 50 m in 4 seconds in the x direction. Link to comment Share on other sites More sharing options...
Salazar Posted September 20, 2012 Author Share Posted September 20, 2012 I appreciate all the input, thanks allot. Link to comment Share on other sites More sharing options...
ACUV Posted September 21, 2012 Share Posted September 21, 2012 Salazar, What is the horizontal distance traveled from hit to catch? Link to comment Share on other sites More sharing options...
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