Homewrok Solution
SOLUTION TO
HOMEWORK PROBLEMS Chapter-4: MOTION IN TWO DIMENSIONS
1 A particle starts from the origin at t = 0 with a velocity of 6.0[pic] m/s and moves in the xy plane with a constant acceleration of (-2.0[pic] + 4.0[pic]) m/s2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?
[pic]
2 At t = 0, a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is given by [pic] = (3.0[pic] - 2.0[pic]) m/s2. At the instant the particle reaches its maximum y coordinate how far is the particle from the origin?
[pic]
3 A particle starts from the origin at t = 0 with a velocity of (16[pic] - 12[pic]) m/s and moves in the xy plane with a constant acceleration of [pic] = (3.0[pic] - 6.0[pic]) m/s2. What is the speed of the particle at t = 3.0 s? v(t) = v(0) +a*t > v(3) = (16i – 12j) + (3i-6j)*3 = 25i – 30j | v(3) | = √(252 + 302) = 39
4 A ball is thrown horizontally from the top of a building 200 m high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?
Yf = Yi + Vit + (1/2) gt2 -200 = 0 + 0 + (-4.9)t2 > t = 6.39 s X = Vixt > Vix = 65/6.39 = 10.17 m/s = Vfx Vfy2 = Viy2 + 2aY > Vfy = √(0 + [2*(-9.8)*200] ) = 62.6 m/s V = √(Vfy2 + Vyx2) = √(62.62 + 10.172) = 63.4 m/s
5 A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?
Find Vx by using Vx = 12.2 cos 53 Find time of flight by: Xf = Xi + Vx t + (1/2) ax t2 (use ax = 0) Use yf = yi + Vy t + (1/2) ay t2 (use ay = -9.8) to calc. yf, which is the height of the building.
HOMEWORK PROBLEMS Chapter-4: MOTION IN TWO DIMENSIONS
1 A particle starts from the origin at t = 0 with a velocity of 6.0[pic] m/s and moves in the xy plane with a constant acceleration of (-2.0[pic] + 4.0[pic]) m/s2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?
[pic]
2 At t = 0, a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is given by [pic] = (3.0[pic] - 2.0[pic]) m/s2. At the instant the particle reaches its maximum y coordinate how far is the particle from the origin?
[pic]
3 A particle starts from the origin at t = 0 with a velocity of (16[pic] - 12[pic]) m/s and moves in the xy plane with a constant acceleration of [pic] = (3.0[pic] - 6.0[pic]) m/s2. What is the speed of the particle at t = 3.0 s? v(t) = v(0) +a*t > v(3) = (16i – 12j) + (3i-6j)*3 = 25i – 30j | v(3) | = √(252 + 302) = 39
4 A ball is thrown horizontally from the top of a building 200 m high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?
Yf = Yi + Vit + (1/2) gt2 -200 = 0 + 0 + (-4.9)t2 > t = 6.39 s X = Vixt > Vix = 65/6.39 = 10.17 m/s = Vfx Vfy2 = Viy2 + 2aY > Vfy = √(0 + [2*(-9.8)*200] ) = 62.6 m/s V = √(Vfy2 + Vyx2) = √(62.62 + 10.172) = 63.4 m/s
5 A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?
Find Vx by using Vx = 12.2 cos 53 Find time of flight by: Xf = Xi + Vx t + (1/2) ax t2 (use ax = 0) Use yf = yi + Vy t + (1/2) ay t2 (use ay = -9.8) to calc. yf, which is the height of the building.