At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40 degrees from horizontal. She leaves the cannon 1m off the ground and lands on the way down in a net 2m off the ground. Determine her horizontal displacement.
Well, it seems like this human cannonball is really looking to fly high and land on a higher note! Let's crunch some numbers and find out her horizontal displacement.
Now, we know that the initial velocity of our cannonball is 15m/s at an angle of 40 degrees from the horizontal. To find the horizontal displacement, we need to resolve this initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity can be found using trigonometry. So, we can say that the horizontal velocity (Vx) is given by Vx = V * cos(theta), where V is the initial velocity and theta is the angle of projection.
Plugging in the values, Vx = 15m/s * cos(40 degrees). Now, I could do some calculations, but hey, let's use a calculator for this!
Calculating... Drumroll, please! Vx comes out to be approximately 11.46 m/s (rounded to two decimal places).
Now that we have the horizontal velocity, we can determine the time it takes for the cannonball to reach its maximum height and then return down to the net. The time of flight can be found using the equation t = (2 * V * sin(theta)) / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the numbers, t = (2 * 15m/s * sin(40 degrees)) / 9.8m/s^2. Let's bring out the calculator once again!
Calculating...voila! The time of flight is approximately 1.89 seconds (rounded to two decimal places).
Now, we know the time and the horizontal velocity. To find the horizontal displacement, we can use the equation d = Vx * t.
Plugging in the values, d = 11.46m/s * 1.89s. "Calculating, calculating" — the horizontal displacement comes out to be approximately 21.65 meters (rounded to two decimal places).
So, our human cannonball will have a horizontal displacement of around 21.65 meters. She's really going to make quite a splash with that cannonball flight! But hey, don't forget to take into account any wind gusts or bird traffic along the way! Safety first, right?
To determine the horizontal displacement of the human cannonball, we can use the horizontal component of the initial velocity and the time it takes for her to reach the net.
Step 1: Find the horizontal component of the initial velocity.
The horizontal component of the initial velocity is given by Vx = V * cos(theta), where V is the magnitude of the initial velocity and theta is the launch angle.
Using the given values, V = 15 m/s and theta = 40 degrees, we can calculate Vx:
Vx = 15 m/s * cos(40 degrees)
Vx ≈ 11.49 m/s
Step 2: Calculate the time it takes for the human cannonball to reach the net.
We can use the vertical motion equation: Δy = Vyi * t + (1/2) * a * t^2, where Δy is the vertical displacement, Vyi is the initial vertical velocity, a is the acceleration, and t is the time.
Since the vertical displacement is given as 2m and the initial vertical velocity is 0 m/s (because the cannonball is launched horizontally), and the acceleration due to gravity is approximately 9.8 m/s^2, we can rearrange the equation to solve for t:
2m = 0 m/s * t + (1/2) * (9.8 m/s^2) * t^2
2m = 4.9 m/s^2 * t^2
t^2 = 2 m / (4.9 m/s^2)
t^2 ≈ 0.41 s^2
t ≈ √0.41 s
t ≈ 0.64 s
Step 3: Calculate the horizontal displacement.
The horizontal displacement (Δx) of an object is given by the equation Δx = Vx * t.
Plugging in the values we calculated, Δx = 11.49 m/s * 0.64 s
Δx ≈ 7.35 m
Therefore, the horizontal displacement of the human cannonball is approximately 7.35 meters.
To determine the horizontal displacement of the human cannonball, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Initial velocity (v₀) = 15 m/s
Launch angle (θ) = 40 degrees
The horizontal component of the initial velocity (v₀x) can be found using the equation:
v₀x = v₀ * cos(θ)
The vertical component of the initial velocity (v₀y) can be found using the equation:
v₀y = v₀ * sin(θ)
Let's calculate v₀x and v₀y:
v₀x = 15 m/s * cos(40°)
v₀x ≈ 11.47 m/s
v₀y = 15 m/s * sin(40°)
v₀y ≈ 9.65 m/s
Next, we need to find the time it takes for the human cannonball to reach its peak height. We can use the vertical component of the initial velocity and the acceleration due to gravity.
The equation for vertical displacement (h) is given by:
h = v₀y * t - 0.5 * g * t²
Since the starting and ending heights are given (1m and 2m respectively), we have:
h = 2 m - 1 m = 1 m
We know that the acceleration due to gravity (g) is approximately 9.8 m/s². Rearranging the equation, we get:
0.5 * g * t² = v₀y * t - h
Plugging in the values, we can solve for time (t):
0.5 * 9.8 m/s² * t² = 9.65 m/s * t - 1 m
0.5 * 9.8 m/s² * t² - 9.65 m/s * t + 1 m = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b² - 4ac)) / (2a)
where,
a = 0.5 * 9.8 m/s²
b = -9.65 m/s
c = 1 m
Solving the quadratic equation, we find two possible solutions for time (t). However, we are only interested in the positive value since time cannot be negative.
Next, we can calculate the horizontal displacement by multiplying the horizontal component of the initial velocity (v₀x) by the time (t):
horizontal displacement = v₀x * t
Plugging in the values for v₀x and t, we can calculate the horizontal displacement.
Vo = 15 m/s @ 40 Deg.
Xo = 15*cos40 = 11.49 m/s.
Yo = 15*sin40 = 9.64 m/s.
Tr = (Yf-Yo)/g,
Tr = (0-9.64) / -9.8 = 0.984 s. = Time
to rise to max. ht.
hmax = ho + (Yf^2-Yo^2)/2g,
hmax = 1 + (0-(9.64)^2) / -19.6=5.74 m
above gnd.
d = Yo*t + 0.5g*t^2 = 5.74 - 2,
0 + 4.9t^2 = 3.74 m,
t^2 = 0.763,
Tf = 0.874 m. = Time to fall to 2 m
above gnd.
Dx = Xo * (Tr + Tf),
Dx=11.49m/s *(0.984+0.874)s =21.4 m. =
Hor. Displacement.