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Answer :
The maximum height reached by the projectile above the top of the cliff is calculated using the projectile's initial vertical velocity and kinematic equations. It is approximately 257 meters.
To find out the maximum height reached by a 58.0-kg projectile above the top of the cliff, we first need to calculate the initial vertical velocity of the projectile and then use that to determine the height it reaches before starting to fall back down. The initial speed of the projectile is 142 m/s and the launch angle is 30.0° above the horizontal.
We can resolve the initial velocity into its vertical component (Vy) using the formula Vy = V * sin(θ), where V is the initial speed and θ is the launch angle. Substituting the given values, we have:
- Vy = 142 m/s * sin(30.0°)
- Vy = 142 m/s * 0.5
- Vy = 71 m/s
Now, to calculate the time (t) it takes to reach maximum height, we know that at the peak of its trajectory, the vertical velocity will be 0 m/s. Applying the kinematic equation V = u + at (where V is the final velocity, u is the initial velocity, a is acceleration due to gravity and is negative in this case, and t is time), we have:
- 0 = 71 m/s - (9.8 m/s² * t)
- t = 71 m/s / 9.8 m/s²
- t ≈ 7.24 seconds
With the time taken to reach the maximum height, we can now calculate the maximum height (H) using the equation H = ut + (1/2)at². Since we are only interested in the height above the initial position, we ignore the height of the cliff and get:
- H = (71 m/s * 7.24 s) - (1/2 * 9.8 m/s² * (7.24 s)²)
- H ≈ 257 meters above the top of the cliff
Therefore, the maximum height reached by the projectile above the top of the cliff is approximately 257 meters.
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