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An attacker at the base of a castle wall 3.65 m high throws a rock straight up with a speed of 7.40 m/s from a height of 1.55 m above the ground.

Determine whether the rock will reach the top of the wall.

Answer :

To analyze the motion of the rock, we can consider the initial and final positions, initial velocity, and acceleration due to gravity.

Given:

Height of the castle wall (h) = 3.65 m
Initial speed of the rock (v₀) = 7.40 m/s
Height above the ground where the rock is thrown (y₀) = 1.55 m
First, let's find the time it takes for the rock to reach its maximum height. We can use the equation:

v = v₀ - gt,

where v is the final velocity (0 m/s), v₀ is the initial velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is the time.

0 = 7.40 - 9.8t,

9.8t = 7.40,

t = 7.40 / 9.8,

t ≈ 0.7551 s.

Next, we can find the maximum height (H) reached by the rock. We can use the equation:

H = y₀ + v₀t - 0.5gt².

H = 1.55 + 7.40 × 0.7551 - 0.5 × 9.8 × (0.7551)²,

H ≈ 5.689 m.

Therefore, the maximum height reached by the rock is approximately 5.689 meters above the ground. Since the height of the castle wall is 3.65 meters, the rock does not reach the top of the wall. It falls back down before reaching that height.

To know more about initial velocity, click here https://brainly.com/question/28395671

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