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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.

(a) Will the rock reach the top of the wall?
(b) If so, what is its speed at the top?

Answer :

Final answer:

The rock reaches the top of the wall and the speed it has when it reaches that height can be calculated by using the kinematic equation for motion under constant acceleration.

Explanation:

This problem involves a physics concept known as Projectile Motion, and it can be solved by using equations related to motion under constant acceleration (also known as kinematic equations).

We can calculate the maximum height of the thrown rock by using the formula: H = (v²)/2g. Where v is the initial launch speed, and g is the acceleration due to gravity. Calculating gives us H = (7.40²)/(2*9.8) = ~2.81m.

Adding the initial height from which the rock was thrown, 1.55m, we get a total height of ~4.36m. So, the rock does reach the top of the castle wall which is 3.65m high.

To calculate the speed with which the rock hits the top of the wall, we use one of the kinematic equations: v²=u²–2gΔh, where: v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and Δh is the change in height. The final speed when it reaches the top of the wall is obtained by substituting the known values into this equation.

Learn more about Projectile Motion here:

https://brainly.com/question/20627626

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