Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Question

contestada

Respuesta :

MechEngineer MechEngineer · Answer 1 · 2020-04-23T04:57:52+02:00

Answer:

a) 8.25 m and 19 m/s

b) 19 m and 24 m/s

c) 32.25 m and 29 m/s

d) 48 m and 34 m/s

e) 66.25 m and 39 m/s

Explanation:

Let gravitational constant g = 10m/s2

Formula for displacement: [tex]s = v_0t + gt^2/2 = 14t + 10t^2/2 = 14t + 5t^2[/tex]

Formula for velocity: v = v_0 + gt = 14 + 10t

a) at t = 0.5s

[tex]s = 14*0.5 + 5*0.5^2 = 7 + 1.25 = 8.25 m[/tex]

v = 14 + 10*0.5 = 14 + 5 = 19 m/s[/tex]

b) at t = 1s

[tex]s = 14*1 + 5*1^2 = 19 m[/tex]

v = 14 + 10*1 = 24 m/s

c) at t = 1.5 s

[tex]s = 14*1.5 + 5*1.5^2 = 21 + 11.25 = 32.25 m[/tex]

v = 14 + 10*1.5 = 29 m/s

d) at t = 2s

[tex]s = 14*2 + 5*2^2 = 28 + 20 = 48 m[/tex]

v = 14 + 10*2 = 34 m/s

e) at t = 2.5s

[tex]s = 14*2.5 + 5*2.5^2 = 35 + 31.25 = 66.25 m[/tex]

v = 14 + 10*2.5 = 39 m/s