The ball is projected up from the ground with speed \[30\text{ }m/sec\]. What is the average velocity for time 0 to 4 sec?
A \[20\text{ }m/sec\]
B \[10\text{ }m/sec\]
C \[15\text{ }m/sec\]
D zero

Hint: Average velocity is equal to the ratio of change in displacement (\[\Delta S\]) to change in the time (\[\Delta T\]) interval. The time interval for which the average velocity is to find is 0 to 4 sec. We need to calculate displacement with the formula such as \[S=ut+\dfrac{1}{2}\text{ }at{}^\text{2}\]. As the ball is projected upward from rest against the force of gravity so, the acceleration of the ball will be the same as acceleration due to gravity (\[g\text{ }=\text{ }10\text{ }m/sec{}^\text{2}\]) but in opposite direction (\[-\text{ }10\text{ }m/sec{}^\text{2}\]).

Formula used:
The displacement formula from equation of motion is,
\[S=ut+\dfrac{1}{2}\text{ }at{}^\text{2}\
where u is initial velocity, t is time interval, a is the acceleration and S is the displacement.

Complete step by step solution:
We need to find the displacement of the ball within the time interval of 0 to 4 sec to find the average velocity of the ball which is projected in an upward direction and as per hint suitable displacement formula is\[S=ut+\dfrac{1}{2}\text{ }at{}^\text{2}\] where u is initial velocity, t is time interval for which displacement is to find and a is acceleration with which ball is moving in an upward direction.

The initial velocity (u) of the ball, while projected upward from the ground, is \[30\text{ }m/sec\]. Change in time (\[\Delta T\]) is equal to final time (\[{{t}_{2}}\]) minus initial time (\[{{t}_{1}}\]) such as\[~\Delta T=\text{ }{{t}_{2}}\text{ }-{{t}_{1}}\], given time interval is from 0 to 4 seconds so the change in time is equal to,
\[\text{ }\Delta \text{T }=\text{ }\left( 4\text{ }\text{ }0 \right)s=4s\]

Acceleration (represented as a) of a body thrown upward is experiencing force against gravity and thus, its acceleration will change to acceleration due to gravity (represented as g) which is equal to \[10\text{ }m/se{{c}^{2}}\]but as the body is moving against gravity so its acceleration is \[-\text{ }10\text{ }m/sec{}^\text{2}\].

Putting all the physical quantities in the formula of displacement (S) such as
\[S=ut+\dfrac{1}{2}\text{ }at{}^\text{2}\]
\[\Rightarrow S\text{ }=\text{ }30\text{ }\left( 4 \right)\text{ }+\dfrac{1}{2}\text{ }\left( -10 \right)\text{ }\left( 4 \right){}^\text{2}\]
\[\Rightarrow S\text{ }=\text{ }120\text{ }-80\]
\[\Rightarrow S\text{ }=\text{ }40\text{ }m\]
Now the average velocity is equal to the ratio of average displacement (total displacement) to the average time (change in time) such as
Average velocity (\[{{V}_{av}}\]) = total displacement / average time
\[{{V}_{av}}=\dfrac{\Delta S}{\Delta T}\]
\[\Rightarrow {{V}_{av}}\text{ }=\text{ }40/4\]
\[\therefore {{V}_{av}}\text{ }=\text{ }10\text{ }m/sec\]

Thus, the correct option is B.

Note: Generally S (displacement) is in two ways, one is \[v{}^\text{2}-\text{ }u{}^\text{2}=\text{ }2aS\]and the other way is such as \[S\text{ }=\text{ }at\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }at{}^\text{2}\]. As per this question suitable formula for finding displacement is \[S\text{ }=\text{ }at\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }at{}^\text{2}\]because we have all the physical quantities (u, t, and t). Whereas in the other formula \[v{}^\text{2}-\text{ }u{}^\text{2}=\text{ }2aS\], we don’t have one variable which is the final velocity (v) of the ball after projection within time interval so we cannot use it.