The question was:
Take g= 10 m/s2. A stone propelled from a catapult with a speed of 50 m/s attains a height of 100 m. Calculate:
a. the time of flight.
b. the angle of projection.
c. the range attained.
Both bots missed it, finally deciding the range was 0 meters! (they assumed the stone was projected vertically.)
However, we all know that
the maximum height is h = v^2 sin^2θ/2g
So that means that
50^2 sin^2θ/20 = 100
sin^2θ = 2000/2500
sin^2θ = 0.8
θ = 63.4°
Now we can answer the questions:
a. the time of flight.
h = 50 sinθ t - 5t^2
when h=0, t = 8.94 s
c. the range attained.
R = v^2 sin2θ/g = 50^2 * 0.8/5 = 200 m
b. To find the angle of projection, we can use the equation for the horizontal range:
R = (v^2 * sin(2θ))/g
where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
Rearranging this equation to solve for θ, we have:
sin(2θ) = (R * g)/(v^2)
2θ = arcsin((R * g)/(v^2))
θ = (1/2) * arcsin((R * g)/(v^2))
Using the given values R = 200 m, v = 50 m/s, and g = 10 m/s^2, we can substitute them into the equation:
θ = (1/2) * arcsin((200 * 10)/(50^2))
θ = (1/2) * arcsin(400/2500)
θ = (1/2) * arcsin(0.16)
θ ≈ 9.5°
So the angle of projection is approximately 9.5°.
Correct!
To find the time of flight, we can use the equation for vertical displacement:
h = v^2 * sin^2θ / 2g
Plugging in the given values:
100 = (50^2) * sin^2θ / (2 * 10)
Rearranging the equation and solving for sin^2θ:
sin^2θ = 2000 / 2500 = 0.8
Taking the square root of both sides:
sinθ ≈ √0.8 ≈ 0.8944
Finding the angle of projection (θ):
θ ≈ sin^(-1)(0.8944) ≈ 63.4°
a. To find the time of flight, we can use the equation for vertical displacement:
h = v * sinθ * t - 0.5 * g * t^2
Plugging in the values:
100 = 50 * sin(63.4°) * t - 0.5 * 10 * t^2
Rearranging the equation and solving for t:
5t^2 - 50 * sin(63.4°) * t + 100 = 0
Using the quadratic formula:
t = [-(-50 * sin(63.4°)) ± √((-50 * sin(63.4°))^2 - 4 * 5 * 100)] / (2 * 5)
Simplifying:
t ≈ 0.893s (ignoring the negative value)
So, the time of flight is approximately 0.893 seconds.
c. To find the range attained, we can use the equation:
R = v^2 * sin(2θ) / g
Plugging in the given values:
R = (50^2) * sin(2 * 63.4°) / 10
Simplifying:
R ≈ 200 meters
So, the range attained is approximately 200 meters.
To solve for the time of flight and range attained, we need to find the angle of projection first. The angle of projection can be determined using the formula:
sin^2θ = (2h / (v^2/g))
Where h is the maximum height (100 m in this case), v is the initial velocity (50 m/s in this case), and g is the acceleration due to gravity (10 m/s^2).
Plugging in the values, we find:
sin^2θ = (2 * 100 / (50^2/10))
= 2000 / 2500
= 0.8
Taking the square root of both sides:
sinθ = sqrt(0.8)
θ = arcsin(sqrt(0.8))
θ ≈ 63.4°
Now that we have the angle of projection, we can answer the questions:
a. The time of flight:
The time of flight can be determined using the equation for vertical motion:
h = v * sinθ * t - (1/2) * g * t^2
Since the stone reaches a height of 100 m, we can set h equal to 100 and solve for t:
100 = 50 * sin(63.4°) * t - (1/2) * 10 * t^2
At its highest point, the stone's height will be 100 m, so we substitute this value for h. Rearranging the equation, we get a quadratic equation:
5t^2 - 50 * sin(63.4°) * t + 100 = 0
Solving this quadratic equation will give us the time of flight. Substituting the coefficients into a quadratic formula solver, we find that t ≈ 8.94 seconds.
b. The angle of projection:
The angle of projection has already been determined earlier. It is approximately 63.4°.
c. The range attained:
The range can be calculated using the equation:
R = v^2 * sin(2θ) / g
Substituting the known values, we get:
R = 50^2 * sin(2 * 63.4°) / 10
= 200 m
Therefore, the stone's range is approximately 200 meters.