Question

At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial velocity of 26 m/s. both of them experience negligible air resistance. how much earlier does the thrown rock strike the ground?

Answer 1

To determine how much earlier the thrown rock strikes the ground, kinematic equations are utilized to solve for the times each rock takes to hit the ground from a 300 m building, with the latter having an initial velocity of 26 m/s. The times are calculated separately then subtracted to find the time difference.

Step-by-step explanation:

The question involves the concepts of projectile motion and falling objects in physics, specifically relating to the time it takes for objects with and without an initial velocity to hit the ground when dropped from a certain height. To solve the problem, one can use kinematic equations to determine the time it takes for each rock to reach the ground and then calculate the difference in time between the two events.

To find the time it takes for the first rock (dropped) to hit the ground, we use the equation d = ½at², where d is the distance the rock falls, a is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time in seconds. For the second rock (thrown downward), the equation d = v₀t + ½at² will be used, where v₀ is the initial velocity of the rock. By solving these equations, we find the time for each scenario and then subtract to find the difference.

Answer 2

t = 5.607 sec

Step-by-step explanation:

s = ut + (1/2)at²

you can figure out the time that the dropped ball to hit the ground quite easily because u = 0 m/s, therefore

s = (1/2)at²

t² =2s/a

t = sqrt(2s/a)

assuming the acceleration from gravity is 9.81 m/s²

t = sqrt(2*300m/9.81 m/s²)

t = 7.8 s

You can use the quadratic equation to find t for the thrown rock and then subtract them. Or you can calculate the final velocity of the thrown rock with v^2 = u^2 + 2as and then use v = u + at to find the time for the thrown rock.

v² = (26 m/s)² + 2 * (9.81 m/s^2 * 300 m)

v² =6562

v = 81 m/s

v = u + at

t = (v - u)/s

t = (81 m/s - 26 m/s)/9.81 m/s^2

t = 5.607 sec