Question

A stone 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)?

Answer 1

(a) How fast is it moving when it reaches 12.0 m?
To determine the velocity as it reaches 12.0 m, we use one of the kinematic equations,
V^2 = Vo^2 + 2gh 
where Vo = 20 m/s. 
           g = -9.8 m/s^2 
           h = 12.0 m. 
V^2 = 20^2 + 2(-9.8)(12.0) 
V^2 = 164.8
V = 12.84 m/s

(b) How long is required to reach this height?
 To determine the maximum height, we use the same equation we used above,
V^2 = Vo^2 + 2gh 
where Vo = 20 m/s. 
           g = -9.8 m/s^2
           V = 0 (since at the maximum height velocity is zero) 
0^2 = 20^2 + 2(-9.8)h 
h = 20.41 m

(c) Why are there two answers for (b)?
There are two answers for b because it would travel a distance up and travel a distance down.

Answer 2

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

Further explanation

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

Formula used

$\large{\boxed{\bold{S=v\:\times\:t}}}$

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

$\large{\boxed{\bold{St=vot+\frac{1}{2}at^2}}}$

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)

• a. Distance reached by stone when thrown = 12 m

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

• b. The stone will reach 12 m when it rises and when it comes back down, we use:

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

• 1. factorization
• 2. perfect squared
• 3. abc formula

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

the formula abc =

$x_{12}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

From equation 4.9t²-20t + 12 = 0

a = 4.9 b = -20 and c = 12

$x_{12}=\frac{20\pm\sqrt{(-20)^2-4.4.9.12} }{2.4.9}$

From the formula, we get

t1 = 0.731

t2 = 3.35

• c. the time obtained at point b is 2 because :

t1 = time taken to reach 12 m high when rising

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

Learn more

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the average velocity

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