A body is projected vertically upward direction. The altitude y is given by \(y=16{{t}^{2}}+4t+5\).The initial speed of the body is

Correct option-1

Concept:-

INSTANTANEOUS SPEED

• The speed of a moving object at a particular instant of time is called instantaneous speed.
• We can define velocity at a particular instant as the limiting value of average velocity when time interval ∆t approaches zero.
• Mathematically we can write the same as follows:

 \(v=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta x}{\Delta t}\) = \(\frac{dx}{dt}\)

• When we calculate average velocity or speed, the selected time interval can be large or small. 
• But when the time interval selected is infinitesimally small then it is known as instantaneous velocity or speed.
• For a moving object, the instantaneous speed at any point is equal to the magnitude of instantaneous velocity but the average speed in a given time interval is always greater or equal to the magnitude of average velocity in that interval.

Calculation:-

Given- 

Expression for distance in an upward direction, i.e. altitude \(y=16{{t}^{2}}+4t+5\)

Instantaneous speed is given as-

 \(v=\frac{dy}{dt}\)

\(\Rightarrow\frac{dy}{dt}=\frac{d\left( 16{{t}^{2}}+4t+5 \right)}{dt}=32t+4\)

\(\therefore v=32t+4\)

Initial speed means speed at t = 0 sec.

So at t = 0 sec, we have

\(v=32\times 0+4=4m/s\)

\(\therefore v=4m/s\)

Hence, Option-1 is correct.