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 11.4 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

velocity

y = 6.92[m]

Vf = 16.305[m/s]

Explanation:

To solve this problem we must use the following kinematics equations, where we must first find the final velocity.

where:

Vi = initial velocity = 11.4[m/s]

Vf = final velocity [m/s]

t = time = 0.5[s]

g = gravity acceleration = 9.81[m/s^2}

Note: the positive sign in the equation above is because the acceleration of gravity goes in the direction of the motion of the rock.

Vf = 11.4 + (9.81*0.5)

Vf = 16.305 [m/s]

Now we can calculate how far the rock goes, using the following equation:

Therefore the Rock will be (70 - 6.92) = 63.07 [m] above the water level

its c i just checked it

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.