Question

When the dried-up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.65 m/s

at an angle of 30.0 ∘ below the horizontal. The seed pod is 0.460 m
above the ground.
How long does it take for the seed to land?
t= ? sec
What horizontal distance does it cover during its flight?
x=? M

Answer 1

To solve this problem, you'll need to break the initial velocity of the seed into its horizontal and vertical components, then use the equations of motion to find the time of flight and horizontal distance.

The initial velocity (v) of the seed is 2.65 m/s. The angle it's launched at (θ) is 30.0 degrees below the horizontal. The height (h) it's launched from is 0.460 m.

First, calculate the horizontal (v_x) and vertical (v_y) components of the velocity. Because the seed is launched downward, the vertical component will be negative:

v_x = v * cos(θ) = 2.65 m/s * cos(30.0) = 2.29 m/s

v_y = v * sin(θ) = -2.65 m/s * sin(30.0) = -1.325 m/s

Next, use the equation of motion to find the time it takes for the seed to hit the ground:

h = v_y * t + 0.5 * g * t^2

Where g is the acceleration due to gravity, which is approximately 9.8 m/s². Solving the equation for t gives:

t = (-v_y - sqrt((v_y)^2 - 4 * 0.5 * g * (-h))) / (2 * 0.5 * g)

Plugging in the values:

t = (1.325 + sqrt((-1.325)^2 - 4 * 0.5 * 9.8 * (-0.460))) / (2 * 0.5 * 9.8)

t = 0.182 seconds

Finally, use the horizontal velocity and time of flight to find the horizontal distance the seed covers:

x = v_x * t = 2.29 m/s * 0.182 s = 0.417 m

So, the seed lands after approximately 0.182 seconds and travels approximately 0.417 meters horizontally.