Question

A ball rolling down a hill was displaced 19.6 m while uniformly accelerating from the rest. If the final velocity was 5.00m/s, what was rate of acceleration?

Answer 1

we are given

A ball rolling down a hill was displaced 19.6 m

so,

$h=19.6m$

the final velocity was 5.00m/s

so, $v=5m/s$

Since, a ball rolling down a hill was displaced

so, we get

$u=0m/s$

we can use kinematic equation formula

$v^2-u^2=2ah$

Here ,

v is final velocity

u is initial velocity

a is acceleration

h is height from which ball is dropped

now, we can plug our values

$5^2-0^2=2a(19.6)$

$2a\left(19.6\right)=5^2$

$2a\left(19.6\right)=25$

$a=0.63776m/s^2$................Answer

Answer 2

Using the equation of Motion, the rate of acceleration of the ball is 0.638 m/s²

Using the 2nd equation of Motion :

• Height, h = 19.6 meters
• Final velocity, v = 5 m/s
• Initial Velocity, u = 0

v² = u² - 2gh (downward motion)

5² = 0 - 2(19.6)a

25 = 0 - 39.2a

39.2a = 25

a = 25/39.2

a = 0.638 m/s²

Therefore, the rate of acceleration is 0.638 m/s²

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