A discus is thrown from a height of 3 feet with an initial velocity of 55 ft/s at an angle of 44° with the horizontal. How long will it take for the discus to reach the ground?
I started off with 55cos(44)=40
then I made an equation
40t-16(t)=2.5 seconds?
I feel like I did this wrong because I didn't add 3 feet anywhere
or do I add 3 in to 55cos(44) giving me an answer of 2.6
Yes, the 3 feet are important.
You can consider the datum (y=0) at 3' from the ground, and the ground is therefore at -3 feet.
The vertical initial velocity, vy, is u*sin(θ), so it gives
vy=55sin(44°)=38.21 approx.
The (vertical) distance travelled is given by:
S=vy*t-(1/2)gt²
where
S=-3 (ground)
vy=55sin(44°), and
g=32.2 ft/s²
Solve for t.
I get -0.08 and 2.45 s.
To solve this problem, let's break down the motion of the discus into horizontal and vertical components.
First, calculate the initial horizontal velocity (Vx). You correctly found Vx = 55 ft/s * cos(44°) = 40 ft/s. This represents the rate at which the discus moves horizontally.
Next, calculate the initial vertical velocity (Vy). Vy is given by Vy = 55 ft/s * sin(44°). However, it's important to take into account the direction of Vy. The discus is thrown upwards, so the vertical velocity should be positive. Therefore, Vy = 55 ft/s * sin(44°) = 36.7 ft/s (approximately).
Now, we can analyze the vertical motion of the discus. It starts from a height of 3 feet and travels upwards before falling back down to the ground. The equation that describes the vertical motion is:
y = yo + Voyt + (1/2)gt^2
where:
y = vertical displacement (distance from the initial height)
yo = initial vertical position (3 feet)
Voy = initial vertical velocity (36.7 ft/s)
g = acceleration due to gravity (-32 ft/s^2, taking downward as the negative direction)
t = time
To determine the time it takes for the discus to reach the ground, we set y = 0 (since the final position is at ground level) and solve for t.
0 = 3 + 36.7t - 16t^2
Now, we solve this quadratic equation for t. There are different methods to solve quadratic equations, such as factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula in this case:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 36.7, and c = 3. Plugging these values into the quadratic formula:
t = (-(36.7) ± √((36.7)^2 - 4(-16)(3))) / (2(-16))
Simplifying the equation further:
t = (-36.7 ± √(1348.89 + 192)) / -32
Now, we calculate the values inside the square root:
t = (-36.7 ± √(1540.89)) / -32
Taking the square root:
t = (-36.7 ± 39.26) / -32
Solving for both positive and negative solutions:
t1 = (-36.7 + 39.26) / -32 ≈ 0.084 seconds
t2 = (-36.7 - 39.26) / -32 ≈ 3.68 seconds
Since time cannot be negative in this context, we discard t2 = 3.68 seconds. Therefore, the discus takes approximately 0.084 seconds to reach the ground.
Note: It appears that you mistakenly used a value of 16 for acceleration due to gravity, which should be -32 ft/s^2. Additionally, the 3 feet initial height is already accounted for in the equation y = yo + Voyt + (1/2)gt^2.
To find the time it takes for the discus to reach the ground, you need to consider the vertical motion of the discus.
First, let's break down the initial velocity of 55 ft/s into its vertical and horizontal components. The vertical component of the velocity is given by 55 sin(44°), and the horizontal component is given by 55 cos(44°).
Next, you need to find the time it takes for the discus to reach the ground. The equation that describes the vertical motion of the discus is:
y = y0 + v0y * t - (1/2) * g * t^2
where:
- y is the position of the discus in the vertical direction (which we want to be 0 when it reaches the ground).
- y0 is the initial height of the discus (3 feet).
- v0y is the vertical component of the initial velocity (55 sin(44°)).
- g is the acceleration due to gravity (-32.2 ft/s^2).
- t is the time it takes for the discus to reach the ground (what we're trying to solve for).
Setting y to 0, y0 to 3 feet, v0y to 55 sin(44°), and g to -32.2 ft/s^2, we can solve for t.
0 = 3 + (55 sin(44°)) * t - (1/2) * (-32.2) * t^2
Simplifying the equation, we have:
0 = 3 + 24.706 * t + 16.1 * t^2
Rearranging the equation, we get:
16.1 * t^2 + 24.706 * t + 3 = 0
Solving this quadratic equation, we find two solutions for t. However, we discard the negative solution since time cannot be negative. The positive solution will give us the time it takes for the discus to reach the ground.
The correct value for the time it takes for the discus to reach the ground is approximately 2.6 seconds.