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)?

(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.

ardni313 ardni313 · Answer 2 · 2019-09-18T04:44:51+02:00

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

[tex]\large{\boxed{\bold{S=v\:\times\:t}}}[/tex]

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

[tex]\large{\boxed{\bold{St=vot+\frac{1}{2}at^2}}}[/tex]

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 =

[tex]x_{12}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

From equation 4.9t²-20t + 12 = 0

a = 4.9 b = -20 and c = 12

[tex]x_{12}=\frac{20\pm\sqrt{(-20)^2-4.4.9.12} }{2.4.9}[/tex]

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