Question

A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. Part A: What is the speed of the rock just before it hits the street?

Part B: How much time elapses from when the rock is thrown until it hits the street?

Answer 1

The speed of the rock just before it hits the street is approximately 20.3 m/s, and the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Step-by-step explanation:

Part A: To determine the speed of the rock just before it hits the street, we can use the concept of conservation of energy. When the rock is at the top of its trajectory, its potential energy is at its maximum and its kinetic energy is at its minimum. When the rock is just before hitting the street, its potential energy is at its minimum (zero) and its kinetic energy is at its maximum. Using the equations for potential energy and kinetic energy, we can calculate the speed of the rock:

Initial potential energy = final potential energy + final kinetic energy

mgh = 1/2m*v^2

where m is the mass of the rock, g is the acceleration due to gravity, h is the height of the building, and v is the final velocity of the rock.

Since the mass of the rock cancels out, we have:

gh = 1/2v^2

Plugging in the values, g = 9.8 m/s^2 and h = 21.0 m, we can solve for v:

v = sqrt(2gh)

v = sqrt(2*9.8*21.0)

v = sqrt(411.6)

v = 20.3 m/s

Therefore, the speed of the rock just before it hits the street is approximately 20.3 m/s.

Part B: To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation for time of flight of a vertically thrown object:

t = 2v/g

where t is the time of flight, v is the initial upward velocity of the rock, and g is the acceleration due to gravity.

Plugging in the values, v = 27.0 m/s and g = 9.8 m/s^2, we can solve for t:

t = 2*27.0/9.8

t = 5.5 s

Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Answer 2

A. 33.77 m/s

B. 6.20 s

Step-by-step explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

`V_(final)^(2)=V_(initial)^(2)+2g(y_(final)-y_(initial))\\V_(f)^(2)=27^(2)-2*9.8(-21-0)=33.77 m/s`

Part B. Using Kinematics expression for distance, time and initial velocity

`y_(final)=y_(initial)+V_(initial)t+(1)/(2) g*t^(2)\\==> -0.5*9.8t^(2)+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s`

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.