Can someone help me with thisproblem please......For the initial velocity i got

Question

Can someone help me with thisproblem please......For the initial velocity i got 5.18 m/s....Inclass we were given the formula x=[(Vo^2)*sin(2*0)]/g.to solve for the horizontal distance......how was this equationderived? Please show diagrams and explain.....TIA


A small steel ball bearing with a mass of 20.0 g is on a shortcompressed spring. When aimed vertically and suddenly released, thespring sends the bearing to a height of 1.37 m. Calculate thehorizontal distance the ball would travel if the same spring wereaimed 31.0 deg from the horizontal.

Explanation / Answer

let the heigh be h, for energy conservation, we have: mgh = (1/2)mv02 =>v0 = 2gh =(2*9.8m/s2*1.37m) = 5.18 m/s when aimed at 31.00, we have: vx = v0 cos(31.0) vy = v0 sin(31.0) the time in air is: t = 2vy / g horizontal distance is: x = vx t = vx*2*vy /g = 2v0cos(31.0)*v0sin(31.0) / g                         = v02 sin(2*31.0) / g =(5.18m/s)2 sin(62.0) / 9.8m/s2 = 2.42 m .

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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