A particle moves in the xy plane with constant acceleration. At time zero, the p

Question

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 1.0 m, y = 2.0 m, and has velocity v = 5.0 m/s i hat bold + -6.0 m/s j hat bold . The acceleration is given by the vector a = 2.0 m/s2 i hat bold + 10 m/s2 j hat bold . Find the velocity vector at t = 2.0 s. Find the position vector at t = 2.0 s. Give the magnitude and direction of the position vector in part (b). A ball launched from ground level lands 2.7 s later on a level field 31 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal.

Explanation / Answer

1)

a) (5+2*2)i + (-6+10*2)j = 9i + 14j

b) x = 1 + 5*2 + 0.5*2*2^2 = 15

y = 2 + -6*2 + 0.5*10*2^2 = 10

c) Magnitude = sqrt(15^2 + 10^2) = 18.02m

Angle = tan^-1 (10/15) = 33.689 deg

2)

2*u*sin(theta)/g = 2.7

u*sin(theta) = 13.23

u^2*sin(2theta)/g = 31

u^2*sin(theta)*cos(theta) = 151.9

u*cos(theta) = 151.9/13.23 = 11.48

tan(theta) = 13.23/11.48

theta = 49 degrees

magnitude = sqrt(13.23^2 + 11.48^2) = 17.516

3)

a) T = sqrt(2h/g) = sqrt(2*11300/9.8) = 48.022s

b) 48.022*860*5/18 = 11471.922m

c) Since relative speed is 0 horizontally, Distance = vertical height = 11.3km

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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