In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 9.80 m, and that the bulls-eye is a horizontal distance of d = 3.3 m from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye? (Neglect air resistance.)

Question

anonymous · Answer

Let's set up an expression for the vertical motion of the pumpkin. We know only gravity acts on the pumpkin$$y(t) = y_0 - {1 \over 2} g t^2$$where $y_0$ is the height of the building. We need to find t such that $y(t) = 0$. $$0 = 9.8 - {1 \over 2} g t^2$$We can solve this for t. 

Now that we know the time of flight, we can find the range. Since no forces act in the horizontal direction, the horizontal displacement can be expressed as$$R = vt$$We know that R = 3.3, and we know t from the previous equation. We can now solve for v.