finding vertical component, magnitude of velocity, angle of impact

Question

An object rolls off a tabletop with a horizontal velocity v0x = 2.5 m/s. The table is at a height y0 = 0.55 m, above the floor. Use a coordinate system with its origin on the floor directly beneath the point where the object rolls off the table, its horizontal x-axis lying directly beneath the object’s trajectory, and its vertical y-axis pointing up.

(b) How far, in meters, does the object land from the edge of the tabletop?

(c)What is the vertical component of velocity, in meters per second, when the object hits the ground? Recall that the positive y-direction is upwards.

(d) What is the magnitude of the velocity (its speed) when it hits the floor?

(e) Find the angle of impact, in degrees below the horizontal.

(f) Enter an expression for the height of the object y, in terms of x, v0x, y0, and g. This expression should not depend on the time.

Answer 1

b) y_o = 1/2gt²; 0.55 = 1/2(9.8)t²; t = sqrt 0.1122 = 0.335 s

d_x = v_oxt = (2.5)(0.335) = 0.838 m

c) v_yf = gt = (-9.8)(0.335) = -3.28 m/s

d) Use Pythagorean Theorem with vertical and horizontal components to get v = sqrt (2.5² + 3.28²) = 4.13 m/s

e) tan θ = 3.28/2.5 = 1.312; θ = 52.7º

f) y = y_o + v_oyt + 1/2gt² but we must get rid of t, so v_ox =x/t and t=x/v_ox

substituting we get: y = y_o + v_oyx/v_ox + 1/2gx²/v²_ox