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°.
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
1 answer