Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th

Question

2 years ago

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Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th

ese seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
If a seed is launched at an angle of 0° with the maximum initial speed, how far from the plant will it land? Ignore air resistance, and assume that the ground is flat. (a) 20 cm; (b) 93 cm; (c) 2.2 m; (d) 4.6 m.

Physics

2 answers:

Paraphin [41] · Answer 1 · 2022-06-11T23:45:37+03:00

Answer:

The Correct answer is C

2.2m

Explanation:

The seeds launched obey a projectile motion.

Since it was launched from an height,

Rmax = (u/g)sqrt(u^2 + 2gH)

For seed1, u = 2.3m/s H = 20cm = 0.2m, g = 9.8m/s.

Rmax = (2.3/9.8)*sqrt(2.3^2 + 2*9.8*0.2)

Rmax = 0.701m

For seed2, u = 4.6m/s H = 20cm = 0.2m, g = 9.8m/s.

Rmax = (4.6/9.8)*sqrt(4.6^2 + 2*9.8*0.2)

Rmax = 2.205m

Therefore, the seed launched with maximum initial speed of 4.6m/s, will land at 2.2m from the plant.

Anton [14] · Answer 2 · 2022-06-11T21:45:28+03:00

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!