Question

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the orange jewelweed (impatiens capensis) has seed pods that explode when lightly touched, launching the seeds as projectiles to disperse them. suppose a seed is launched at 1.70 m/s from a height of 1.10 m. assume air resistance is negligible and that the seed follows a clear path to the ground.

if the seed is launched horizontally, at what horizontal distance from the seed pod does the seed hit the ground?
\boxed{} m

Explanation:

Step1: Analyze vertical motion

The vertical motion is free - fall. The initial vertical velocity \(v_{0y} = 0\ m/s\), the vertical displacement \(y=- 1.10\ m\) (taking downwards as negative), and the acceleration \(a = g=- 9.8\ m/s^{2}\). We use the equation \(y=v_{0y}t+\frac{1}{2}at^{2}\). Substituting the values, we get \(- 1.10=0\times t+\frac{1}{2}\times(-9.8)t^{2}\), which simplifies to \(4.9t^{2}=1.10\).

Step2: Solve for time \(t\)

From \(4.9t^{2}=1.10\), we can solve for \(t\) as \(t = \sqrt{\frac{1.10}{4.9}}\approx\sqrt{0.2245}\approx0.474\ s\).

Step3: Analyze horizontal motion

The horizontal motion is uniform motion with \(v_{x}=1.70\ m/s\) (constant velocity as there is no air resistance). The horizontal displacement \(x = v_{x}t\).

Step4: Calculate horizontal distance

Substitute \(v_{x}=1.70\ m/s\) and \(t = 0.474\ s\) into the formula \(x = v_{x}t\), we get \(x=1.70\times0.474 = 0.798\ m\).

Answer:

0.798 (approximate value, more precise calculation may vary slightly)