Question

Assume air resistance is negligible unless otherwise stated.

Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial velocity of 13.2 m/s from the Verrazano Narrows bridge in New York City. The roadway of this bridge is 70.0 m above the water. (Enter the magnitudes.)

(a)

0.500 s

displacement m

velocity m/s

(b)

1.00 s

displacement m

velocity m/s

(c)

1.50 s

displacement m

velocity m/s

(d)

2.00 s

displacement m

velocity m/s

(e)

2.50 s

displacement m

velocity m/s

Answer 1

Given:

• v_i = 13.2 m/s
• x_tot = 70 m
• t₁ = 0.5 s
• t₂ = 1.0 s
• t₃ = 1.5 s
• t₄ = 2.0 s
• t₅ = 2.5 s

According to kinematic equation of motion and given information conclude that v is given by

v = v_i + gt ____(1)

Also get the equation of displacement

x = v_it + 0.5 gt² ____(2)

Apply the previous notation on every time.

(a) For t₁ = 0.5 s

v₁ = v_i + gt = 13.2 + (9.8)(0.5) = 18.1 m/s

x₁ = v_it + 0.5 gt² = (13.2*0.5) + (0.5)(9.8)(0.5²) = 7.825 m

(b) For  t₂ = 1.0 s

v₂ = v_i + gt = 13.2 + (9.8)(1.0) = 23 m/s

x₂ = v_it + 0.5 gt² = (13.2*1.0) + (0.5)(9.8)(1.0) = 18.1 m

(c) For  t₃ = 1.5 s

v₃ = v_i + gt = 13.2 + (9.8)(1.5) = 23 m/s

x₃ = v_it + 0.5 gt² = (13.2*1.5) + (0.5)(9.8)(1.5*1.5) = 30.825 m

(d) For  t₄ = 2.0 s

v₄ = v_i + gt = 13.2 + (9.8)(2.0) = 32.8 m/s

x₄ = v_it + 0.5 gt² = (13.2*2.0) + (0.5)(9.8)(2.0*2.0) = 46 m

(e) For  t₅ = 2.5 s

v₅ = v_i + gt = 13.2 + (9.8)(2.5) = 37.7 m/s

x₅ = v_it + 0.5 gt² = (13.2*2.5) + (0.5)(9.8)(2.5*2.5) = 63.63 m