Question

A ball is thrown straight down with a velocity of 12 m/s; what will be its velocity 2.0 s after being released? Show All Work!

Answer 1

Answer:

v= 32 m/s

Explanation:

Given: u= 12 m/s

a= 10 m/s

t= 2 sec

v= ?

we know that,

v= u+at (1st equation of motion)

v= 12+(10*2)

v= 12+20

v= 32 m/s.

Answer 2

The velocity of the ball thrown straight down with a velocity of 12m/s after 2 sec = 32m/s.

Given:

• Initial velocity of ball = 12m/s
• Time = 2 sec

To find: Velocity of the ball after being released at 2 sec.

Step by step explanation:

According to the first equation od motion,
v = u + a * t
where, v stands for the velocity
           u is the initial velocity of the object
           a is the acceleration of the abject
    and t is the time at which the velocity needs to be calculated.

Here, u = 12m/s
a = acceleration due to gravity as the ball is falling down = 10m/s
t = 2 sec
⇒ v = u + a * t
⇒ v = 12 + 10 * 2
⇒ v = 12 + 20
⇒ v = 32 m/s

∴ The velocity of the ball when dropped from a height with an initial velocity 12m/s after 2 sec = 32m/s.

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