Question

A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of \[19.6m\], he opens the parachute and declares at the rate of \[1m{s^{ - 2}}\]. If he reaches the ground with a speed of \[4.6m{s^{ - 1}}\], how long was he in air \[?\]
A. \[10\] s
B. \[12\] s
C. \[15\] s
D. \[17\] s

Answer 1

Hint: To solve this question, we have to learn the formulas about acceleration, velocity and time. And also we have to know what acceleration is, velocity. We know that, Acceleration is the rate of change of velocity and velocity is the rate of change of speed or speed which is necessary for an object to move in a particular direction.

Complete step by step answer:
Let us consider, the time before opening the parachute is \[{t_1}\]. We are considering h is the height or the distance. U is the initial velocity g is the gravitational acceleration. \[h = ut + \dfrac{1}{2}g{t^2}\], we are going to use this equation to solve this problem. Now after putting the value we get, \[19.6 = 0 + \dfrac{1}{2}9.8{t_1}^2\]
Or, \[{t_1}\]\[ = \]\[2\] s
Now, we are taking \[{v_1}\] as velocity after falling through \[19.6\] m
Now, we know that, \[{v^2} = {u^2} + 2gh\]
Using this equation, after putting the values , we can write,
${v_1}^2 = 0 + 2 \times 9.8 \times 19.6 \\
\Rightarrow{v_1} = 19.6m/s $
Now, we are going to consider, \[{t_2}\] is the time after opening the parachute
We know that,
\[v = u + at\]
Now putting the values we get,
$4.6 = 19.6 - 1 \times {t_2} \\
\Rightarrow{t_2} = 15s$
So, the total time will be \[2 + 15 = 17s\].

So, the write option would be option D.

Note: We can get confused between velocity and acceleration. Speed is equal to distance upon time. and velocity is equal to change of speed upon time. and we know that acceleration is equal to change in velocity by change in time.