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. Will the rock reach the top of the wall?

By using the physics principles of kinematics, we can calculate that the rock, when thrown with an initial velocity of 7.4 m/s from a height of 1.55m, will indeed reach the top of the 3.65m high wall.

Explanation:

This problem is a classical use of basic kinematics in physics, more specifically, the equation of motion for freely falling objects. Firstly, we must account for the initial height of the rock when it is released. The height of the wall relative to the thrower is not 3.65m, but 3.65m - 1.55m = 2.1m.

Now, we can use the equation v² = u² - 2gs, where v is the final velocity, u is the initial velocity (7.4 m/s), g is the acceleration due to gravity (approx. -9.8 m/s²), and s is the distance to be covered (2.1m). We assume that the rock reaches the top of the wall with zero velocity. Solving this equation gives us: 0 = 7.4² - 2*9.8*2.1. After solving this, we find that the left-hand side is more than the right-hand side which signifies that the rock can indeed reach the top of the wall.

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