Astone is thrown vertically upward with a speed of 20.0 m/s. (a) how fast is it moving when it reaches 12.0 m? (b) how long is required to reach this height? (c) why are there two answers for (b)?

a. The speed when  it reaches 12.0 m = Vt = 12,837 m/s

b. The time to reach 12.0 m height

t1 = 0.731

t2 = 3.35

c. There two answers for (b) because there is up and down time

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

S = distance = m

v = speed = m / s

t = time = seconds

Straight motion changes regularly are the straight motion of objects that have a fixed acceleration

Formula used

V = vo + at

Vt² = vo² + 2as

St = distance on t

vo = initial speed

vt = speed on t

a = acceleration

The vertical upward motion is the part of irregularly changing motion where the acceleration is fixed and the magnitude equal to the acceleration due to gravity (a = g)

A stone is thrown vertically upward with a speed of 20.0 m / s, this is Vo (initial speed)

Then the speed:

Vt² = Vo² - 2gs (sign - indicates opposite to Earth's gravity)

Vt² = 20² - 2.9.8.12

Vt² = 164.8

Vt = 12,837 m / s

St = Vot - 1/2gt²

12 = 20.t-1/2.9.8t²

4.9t²-20t + 12 = 0

There are 3 solutions for quadratic equations like this

We use the abc formula, because in ways 1 and 2 it is rather difficult

the formula abc =

From equation 4.9t²-20t + 12 = 0

a = 4.9 b = -20 and c = 12

From the formula, we get

t1 = 0.731

t2 = 3.35

t1 = time taken to reach 12 m high when rising

t2 = the time taken to reach the height of 12 m when descending

The distance of the elevator

link

resultant velocity

link

the average velocity

link

Keywords: vertical motion, gravitational acceleration, stone, fixed acceleration

they should be similar or the same

an electric circuit