An object is launched at a speed of 15 m/s. If it rises to a height of 2 m, at what angle was it launched?
Vo = 15 m/s.
Yf^2 = Yo^2 + 2g*h.
Yf = 0.
g = -9.8 m/s^2.
h = 2 m.
Yo = ?.
sin A = Yo/Vo.
A = ?
To determine the angle at which the object was launched, you can start by using the formula for the maximum height reached by a projectile:
h = (v^2 * sin^2(θ))/(2 * g)
Where:
h = maximum height reached
v = initial velocity of the object
θ = launch angle
g = acceleration due to gravity (approximately 9.8 m/s^2)
Given that the object reaches a height of 2 m and an initial velocity of 15 m/s, we can substitute these values into the equation:
2 = (15^2 * sin^2(θ))/(2 * 9.8)
Simplifying the equation:
2 = (225 * sin^2(θ))/ 19.6
Now, we can isolate sin^2(θ) by multiplying both sides by 19.6 and dividing by 225:
sin^2(θ) = (2 * 19.6) / 225
sin^2(θ) ≈ 0.175
To find sin(θ), you can take the square root of both sides of the equation:
sin(θ) ≈ √0.175
sin(θ) ≈ 0.418
Lastly, to find the angle θ, you can take the inverse sine (sin^(-1)) of 0.418 using a scientific calculator or trigonometric table:
θ ≈ sin^(-1)(0.418)
θ ≈ 24.2 degrees
Therefore, the object was launched at an angle of approximately 24.2 degrees.