Question

A stone is thrown vertically upward with a velocity of 98m/s. Its velocity will be zero after \[(g=9.8m/{{s}^{2}})\]:
A) 14s
B) 10s
C) 20s
D) 5s

Answer 1

Hint: Here a stone is thrown upward with velocity 98m/s and we have to find the time at which its velocity becomes zero. Now as the stone is thrown upward after reaching a certain distance it will fall back to the ground due to gravitation and the point at which it starts to fall downward or the maximum height till which it can go is the point where it has zero velocity. So we have to find the time taken by stone to reach the maximum height when its initial velocity is 98m/s and final velocity is zero.

Formula used:
\[\begin{align}
  & {{v}^{2}}={{u}^{2}}+2as \\
 & s=ut+\dfrac{1}{2}a{{t}^{2}} \\
\end{align}\]

Complete step by step answer:
Here we have to find the time at which final velocity of the stone is zero when it is thrown vertically upward. And we know that any object which is thrown in upward direction after travelling a certain distance falls back to the surface or ground. That particular distance at which it starts following back is where its velocity is zero.
Now the acceleration with which the stone goes up or fall back can we given as the acceleration due to gravity as in first case that is going up it need to having same or more acceleration to go against the gravitational force and in second case of falling back it is due to the gravitational pull so it will be having the acceleration due to gravitational effect. A simple representation is shown below

Now the equation relating initial velocity u, final velocity v, acceleration a and distance s is given s
\[{{v}^{2}}={{u}^{2}}+2as\]
By using the above equation let us find the maximum height or distance at which the stone can reach, as we have all the other parameters i.e. v, u and a is given. Here \[v=0,u=98m/s\text{ and }a=g=9.8m/s\]. Substituting all these value in above equation we get
\[\begin{align}
  & {{\left( 0 \right)}^{2}}={{\left( 98 \right)}^{2}}+2\left( 9.8 \right)s \\
 & \Rightarrow 0=9604+19.6s \\
 & \Rightarrow 19.6s=-9604 \\
 & \Rightarrow s=-\dfrac{9604}{19.6} \\
 & \Rightarrow s=-490m \\
\end{align}\]
Here s shows displacement and so it represents that velocity is decreasing with height. It also shows that acceleration g is in opposite to the displacement.
To find time taken to attain zero velocity let us consider the case when it falls back to the ground. When it falls back, distance would be the same and acceleration will also be the same. Although its initial velocity will be zero while falling backwards.
Now the formula for distance in terms of acceleration, velocity and time is
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
Substituting value of initial velocity for second case zero and acceleration \[g=9.8m/s\] and distance s as 490m (because while falling backwards its velocity will increase so no need for negative sign), in above equation we get
\[\begin{align}
  & 490=\left( 0 \right)t+\dfrac{1}{2}\left( 9.8 \right){{t}^{2}} \\
 & \Rightarrow 490=4.9{{t}^{2}} \\
 & \Rightarrow {{t}^{2}}=\dfrac{490}{4.9} \\
 & \Rightarrow {{t}^{2}}=100 \\
 & \Rightarrow t=\sqrt{100}=10s \\
\end{align}\]
Hence stones velocity will be zero after 10 seconds.

So, the correct answer is “Option B”.

Note: We could have used a distance formula for the first case to find time but in that case we would have got a quadratic equation and the solution becomes lengthy. Although the result from the first case and the second case would be the same as the distance and acceleration is the same. Note that the final velocity of the first case is the same as the initial velocity for the second case.