Answer :
Final answer:
Given the projectile motion problem, the height of the rock 1.57 seconds before it hits the ground is calculated to be approximately 109.29 meters
Explanation:
This is a problem of kinematics, where we need to compute the height of a rock given information about its initial velocity, angle of projection, and time before hitting the ground. The key equation here is H = V * t - 0.5 * g * t2.
The first step is finding the overall time of flight, which is (2 * V * sinθ) / g; plugging the values in (2 * 26.4 * sin58.9) / 9.8 = 5.03s. So, 1.57 seconds before hitting the ground would be at t = 5.03 - 1.57 = 3.46s.
To find the vertical height at this moment, plug the values into the equation: H = 26.4 * sin58.9 * 3.46 - 0.5 * 9.8 * 3.462 = 48.39m. Adding the height of the cliff, we get 60.9 + 48.39 = 109.29m. So, the rock is 109.29 meters above the ground 1.57 seconds before it hits the ground.
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