An attacker at the base of a castle wall 3.53 m high throws a rock straight up with speed 6.70 m/s from a height of 1.54 m above the ground. what is its speed at the top?

Question

parthkohli · Answer

$$v^2 = u^2 - 2gs$$Use this equation.

parthkohli · Answer

$$u = \text{initial velocity }=6.7m/s$$$$s = \text{distance covered }=3.53 - 1.54 \text{meters}$$$$v \text{ is to be calculated}$$

anonymous · Answer

i think you should draw it first; then derive the formula

looks like a calculus problem

parthkohli · Answer

$$g = {9.8} {m \over s^2}$$

anonymous · Answer

question, is the top referring to the wall or the maximum height the rock reaches? and the rock starts at the bottom of the wall?

anonymous · Answer

velocity equations follow as:
$$a x^{2} + bx + c = d$$ where a is the acceleration, b is velocity , and c is position of the rock. d is the final position of the rock.