In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.
Let us list out what we know from the question so that we can decide on the equation to be used.
We know that Initial Y Velocity
= 8.4 m/s
Acceleration in the Y direction 
= -9.8 m/
, since the acceleration due to gravity points in the downward direction.
Final Y Velocity 
= 0 because at the highest point in its path, an object comes to rest momentarily before falling down.
Time taken t = ?
From the list above, it is easy to see that the equation that best suits our purpose here is 
Plugging in the numbers, we get 0 = 8.4 - (9.8)t
Solving for t, we get t = 0.857 s
Therefore, the snowball takes 0.86 seconds to reach its highest point.
There are a few ways to do this- unfortunately different fields are better at it than others! Medical research is generally pretty good, some other fields likewise very good, some not as much.
Basically, though, what they do is use standadisation- they agree on the terminology, units of data, statistical measures, and so forth, that will be used in that scientific field. As much as possible, every scientist in the field uses those standards so everyone working in the field should recognise it.
For instance, in clinical trials, there is very good agreement worldwide on what the different metrics we use are- e.g. in cancer research, we usually want to know the 5-year survival rate (meaning the percentage of patients still alive 5 years after diagnosis). So anyone with the right training should be able to pick up a clinical trial report and understand what the results are and what the report is saying.
Helping verbs come before the main verb, the main verb in the sentence is repair therefor the helping verb would be could.
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
