The equation that solves a problem is 6.4 m = 20 m + 3.0 m/s(2.0 s) - 4.9 m/s2(2

Question

The equation that solves a problem is 6.4 m = 20 m + 3.0 m/s(2.0 s) - 4.9 m/s2(2.0 s)2. The problem is; How far above its initial position does a rock travel in 2.0 s when thrown up from a point 40 m above the ground? How far below its Initial position does a rock travel in 2.0 s when thrown up from a point 40 m above the ground? What is the position relative to the ground of a rock thrown up at 3.0 m/s from a roof 20 m above the ground 2.0 s after it is released? What is the change in position relative to the ground of a rock thrown up at 3.0 m/s from a roof 20 m above the ground 2.0 s after it is released? What is the position relative to the ground of a rock thrown up at 3.0 m/s from a roof 20 m above the ground if its maximum height is 33.6 m?

Explanation / Answer

The given equation is : 6.4 m = 20 m + 3.0 m/s (2.0 s) - 4.9 m/s^2(2.0 s)^2 Rearrange the equation as: 6.4 m = 20 m + 3.0 m/s (2.0 s) - 0.5*9.8 m/s^2(2.0 s)^2 Compare the above equation with the equation of motion of constant acceleration system s = s0 + v0*t + 0.5*g*t^2 Where s-s0 is the displacement v0 is the initial velocity g is acveeleration due to gravity t is the time Form above equations, we can conclude that this will be a solution of "option C" What is the position relative to the ground of a rock thrown up at 3.0 m/s from aroof 20 m above the ground 2.0 s after it is released?

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.