Question

In Denver children bring their old jack o lanterns to the top of a tower and compete for accuracy I hitting a target on the ground. suppose that the tower is 9.0m high and that the bulls eye is a horizontal distance of 3.5m from the launch point. if the pumpkin is thrown horizontally what is the launch speed needed to hit the bulls eye?

Answer 1

Givens.

• The height of the tower is h = 9.0 m.

,

• The bulls-eye is 3.5 m away from the tower, x = 3.5 m.

Before we find the needed speed to hit the bulls-eye, we need to find the time. Use a formula that includes height, gravity, and time.

$y=-\frac{1}{2}gt^2$

This formula does not show the initial velocity because is null. Use the given magnitudes and solve for t.

$\begin{gathered} -9m=-\frac{1}{2}\cdot9.8\cdot\frac{m}{s^2}\cdot t^2 \\ -9m=-4.9\cdot\frac{m}{s^2}\cdot t^2 \\ t^2=\frac{9m}{4.9\cdot\frac{m}{s^2}} \\ t\approx1.36\sec \end{gathered}$

Find the final velocity. In this case, use the formula for a constant motion because the pumpkin is thrown horizontally, and the horizontal motion is constant.

$\begin{gathered} x=v_x\cdot t \\ v_x=\frac{x}{t}=\frac{3.5m}{1.36s} \\ v_x=2.57\cdot\frac{m}{s} \end{gathered}$

Therefore, the needed speed to hit the bulls-eye is 2.57 m/s.