Projectile Motion

[Salazar]

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.

[swansont]

What do you have so far for a solution?

[Salazar]

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)?

[ACUV]

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

[swansont]

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.

[Salazar]

I appreciate all the input, thanks allot.

[ACUV]

Salazar,

 

What is the horizontal distance traveled from hit to catch?