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

The equation that relates distance, velocities, acceleration, and time is,
                   d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and 
g is the acceleration due to gravity (equal to 9.8 m/s²)

(1) Dropped rock,
                  (3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
                (3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s

The difference between the tim,
        difference = 24.73 s - 5.61 s
          difference = 19.12 s

ANSWER: 19.12 s

Answer 2

Answer:

t = 5.607 sec

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