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Answer :
Final answer:
D 26.7 m/s,36.1 degrees above horizontal. To find the launch velocity of the projectile, we can use the kinematic equations of motion. We can solve for the time it takes for the projectile to reach its maximum height and use that to calculate the launch velocity. The launch velocity of the projectile is approximately 26.7 m/s, 36.1 degrees above horizontal.
Explanation:
To find the launch velocity of the projectile, we can use the kinematic equations of motion. We know that the projectile reaches a maximum height, which means its vertical velocity is zero at that point. We can use this information to find the time it takes for the projectile to reach its maximum height. Using the time and the horizontal distance traveled, we can calculate the launch velocity.
Let's assume the launch angle is θ and the launch velocity is V0. At the maximum height, the vertical velocity is given by:
vy = V0sin(θ) - gt = 0
Using this equation, we can solve for the time it takes for the projectile to reach its maximum height:
t = V0sin(θ)/g
Next, we can calculate the horizontal distance the projectile travels:
d = V0cos(θ) * t
Plugging in the known values, we get:
46.8 m = V0cos(θ) * V0sin(θ)/g
Using trigonometric identities, we can rewrite the equation as:
46.8 m = V02 * sin(2θ)/g
Since the launch angle is not given, we have to find it using the equation:
tan(θ) = Vy/Vx
Using the known values, we get:
tan(θ) = H/V0
Solving for θ, we find:
θ = atan(H/V0)
Finally, we can substitute the calculated θ into the equation for the launch velocity:
V0 = sqrt((g * d)/sin(2θ))
Plugging in the given values, we find that the launch velocity of the projectile is approximately 26.7 m/s, 36.1 degrees above horizontal.
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