A student sits atop a platform a distance above the ground. He throws a pingpong

Question

A student sits atop a platform a distance above the ground. He throws a pingpong ball horizontally with a speed . However, a wind blowing parallel to the ground gives the pingpong ball a constant horizontal acceleration with magnitude . This results in the pingpong ball reaching the ground directly under the student.

Determine the height in terms of , , and . This is a somewhat unphysical problem in that you can ignore the effect of air resistance on the vertical motion, but clearly the wind has a big affect on the horizontal motion.


Hint: the pingpong ball moves with constant acceleration in both x and y. You cannot use the projectile motion formulas, but must use the equations of motion with both a_x and a_y=-g. The fall time of the pingpong ball is determined by the y-component equations with -g. The inital velocity only has an x-component, and the displacement in x is zero. Solve for time to reach x_final=0 and equate to the fall time.

Explanation / Answer

* In horizontal direction the position of the ball is
x=v*t –0.5a*t^2;
since on the ground x=0, then 0=v*t –0.2a*t^2, hence
time in free fall is t=2v/a;
* In free fall: depth of fall is y=0.5g*t^2;
since on the ground y=h, then
h=0.5g*t^2 = 0.5g*(2v/a)^2

=2g(v/a)^2

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