When the dried up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.2 m/s at an angle of 60.0° above the horizontal.
If the seed pod is 0.8 m above the ground, how long does it take for the seed to land? (Neglect air resistance.)
What horizontal distance does it cover during its flight?
Vo = 2.2m/s @ 60 deg.
Vo(v) = 2.2sin60 = 1.91m/s.
Vf^2 = Vo^2 + 2gd,
d = (Vf^2-Vo^2) / 2a,
d = (0-(1.91)^2 / -19.6 = 0.186m,up.
Vf = Vo + gt,
t = (Vf - Vo) / a,
t = (0 - 1.91 / -9.8 = 0.195s,up.
d = 0.8 + 0.186 = 0.986m above ground.
d = Vo + 0.5gt^2 = 0.986m,
0 + 0.5*9.8t^2 = 0.986,
4.9t^2 = 0.986,
t^2 = 0.20,
t = 0.447s, down.
T = t(up) + t(down) = 0.195 + 0.447 =
0.64s to land.
Vo(h) = 2.2cos60 = 1.1m/s.
d(h) = 1.1m/s * 0.64s. = 0.704m = hor
distance.
Well, aren't you just bursting with questions? Let's calculate it, shall we?
To find the time it takes for the seed to land, we can use the vertical motion equation:
y = y₀ + v₀y * t - (1/2) * g * t²
Where:
y = final vertical displacement (0 m since it lands on the ground)
y₀ = initial vertical position (0.8 m)
v₀y = initial vertical velocity (2.2 m/s * sin(60°))
g = acceleration due to gravity (9.8 m/s²)
t = time
Plugging in the values, we get:
0 = 0.8 + (2.2 * sin(60°)) * t - 0.5 * 9.8 * t²
Solving this quadratic equation will give us the time it takes for the seed to land. Let me grab my trusty calculator...
(Seconds later...)
After some number crunching, the time comes out to be approximately 0.438 seconds. So it takes the seed about 0.438 seconds to land.
Now, to find the horizontal distance covered during flight, we can use the equation:
x = v₀x * t
Where:
x = horizontal distance
v₀x = initial horizontal velocity (2.2 m/s * cos(60°))
t = time (0.438 s)
Calculating again...
(Drumroll please...)
The horizontal distance covered during flight is approximately 0.955 meters. So it travels about 0.955 meters horizontally.
And there you have it! The seed's journey from being a flying burst of excitement to tumbling down to the ground. Seed-iously, it's quite a feat!
To find the time it takes for the seed to land, we can use the equation of motion in the vertical direction:
h = vit + (1/2)gt^2
Where:
h = initial height above the ground (0.8 m)
vi = initial vertical velocity (v*sinθ)
g = acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Using the given values:
vi = 2.2 m/s * sin(60°) ≈ 1.90 m/s
g = -9.8 m/s^2
h = 0.8 m
0.8 = (1.90)t + (1/2)(-9.8)t^2
Rearranging the equation:
-4.9t^2 + 1.90t + 0.8 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
a = -4.9
b = 1.90
c = 0.8
Plugging the values into the quadratic formula:
t = (-1.90 ± √(1.90^2 - 4(-4.9)(0.8)))/(2(-4.9))
Calculating the values inside the square root:
t = (-1.90 ± √(3.61 + 15.36))/(2(-4.9))
t = (-1.90 ± √(19.97))/(2(-4.9))
t = (-1.90 ± √19.97)/(-9.8)
Calculating the square root:
t = (-1.90 ± 4.47)/(-9.8)
Using the positive root (since time cannot be negative):
t = (-1.90 + 4.47)/(-9.8)
t ≈ 0.251 s
Therefore, it takes approximately 0.251 seconds for the seed to land.
To find the horizontal distance covered during its flight, we can use the equation of motion in the horizontal direction:
d = v*cosθ * t
Where:
d = horizontal distance
v = initial velocity (2.2 m/s)
θ = angle above the horizontal (60°)
t = time of flight (0.251 s)
Plugging the values into the equation:
d = 2.2 m/s * cos(60°) * 0.251 s
d ≈ 0.962 m
Therefore, the seed covers approximately 0.962 meters horizontally during its flight.
To find the time it takes for the seed to land, we can use the vertical component of motion. The initial vertical velocity (Vy) can be found using the given initial velocity (2.2 m/s) and the angle of projection (60.0°).
Vy = V * sin(theta)
Vy = 2.2 m/s * sin(60.0°)
Vy = 2.2 m/s * (√3 / 2)
Vy = 1.9 m/s
Next, we can use the vertical motion equation to find the time of flight (t):
h = Vy * t + (1/2) * g * t^2
Where h is the initial vertical position (0.8 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
0.8 m = 1.9 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Rearranging the equation and setting it equal to zero:
(1/2) * 9.8 m/s^2 * t^2 + 1.9 m/s * t - 0.8 m = 0
Now we can solve this quadratic equation for time (t) using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = (1/2) * 9.8 m/s^2, b = 1.9 m/s, and c = -0.8 m.
t = [ -1.9 ± √(1.9^2 - 4 * (1/2) * 9.8 * (-0.8)) ] / (2 * (1/2) * 9.8)
After solving for t, we can ignore the negative solution because time cannot be negative in this context. The positive solution will give us the time it takes for the seed to land.
To find the horizontal distance covered during the flight, we can use the equation:
d = Vx * t
Where Vx is the horizontal component of the initial velocity. We can find Vx using the given initial velocity and the angle of projection:
Vx = V * cos(theta)
Vx = 2.2 m/s * cos(60.0°)
Vx = 2.2 m/s * (1/2)
Vx = 1.1 m/s
Now we can substitute the value of t into the equation to find the horizontal distance (d) covered during the flight.
d = 1.1 m/s * t
Once we have the value of t, we can calculate the horizontal distance.