The equation of motion of a stone, thrown vertically upwards is $s = ut - 6.3{t^2}$, where the units of $s$ and $t$ are in $cm$ and $\sec $. If the stone reaches maximum height in $3{\text{ }}\sec $ then, $u = $?
(i) $18.9{\text{ }}\dfrac{{cm}}{{\sec }}$ (ii) $12.6{\text{ }}\dfrac{{cm}}{{\sec }}$ (iii) ${\text{37}}{\text{.8 }}\dfrac{{cm}}{{\sec }}$ (iv) None of these
Answer
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Hint:When a ball is thrown upwards it eventually attains a maximum height and then comes back down. We will solve this problem by analysing the given equation. Then using 1st Motion’s equation from a concept that at maximum height the velocity of a body is zero.
Formula used: (A) $v = u + at$
where $v = $ final velocity, $u = $initial velocity, $a = $ acceleration and $t = $ time
(B) $s = ut + \dfrac{1}{2}a{t^2}$
Where $u = $initial velocity, $t = $ time, $a = $ acceleration and $s = $ distance covered.
Complete step-by-step solution:
The given expression of motion is 2nd Motion equation as,
$s = ut - 6.3{t^2} - - - - - \left( 1 \right)$
2nd equation of motion is,
$s = ut + \dfrac{1}{2}a{t^2} - - - - - \left( 2 \right)$
Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$a = - 12.6$
According to the question the ball attains its maximum height in $3{\text{ }}\sec $.
When a ball attains a maximum height then its final velocity becomes $0$.
Substituting the values in 1st Motion’s equation, $v = u + at$ where $v = $ final velocity, $u = $initial velocity, $a = $ acceleration and $t = $ time, we get
$0 = u - 12.6 \times 3$
$ \Rightarrow u = 37.8$
So, the initial velocity $u = 37.8{\text{ }}\dfrac{m}{s}$.
Alternative method:
We can also find $u$ from the derivative method. When the ball attains a maximum height then,
$\dfrac{{ds}}{{dt}} = 0$ from maximum property.
So, $u - 12.6 \times 3 = 0$
$ \Rightarrow u = 37.8$
Note:Students tend to make errors while solving these types of questions because they fail to apply the formula correctly, as well as to use the correct sign based on the direction of motion of the object. So, they should make sure to understand the problem's scenario clearly and then solve it without any calculation error.
Formula used: (A) $v = u + at$
where $v = $ final velocity, $u = $initial velocity, $a = $ acceleration and $t = $ time
(B) $s = ut + \dfrac{1}{2}a{t^2}$
Where $u = $initial velocity, $t = $ time, $a = $ acceleration and $s = $ distance covered.
Complete step-by-step solution:
The given expression of motion is 2nd Motion equation as,
$s = ut - 6.3{t^2} - - - - - \left( 1 \right)$
2nd equation of motion is,
$s = ut + \dfrac{1}{2}a{t^2} - - - - - \left( 2 \right)$
Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$a = - 12.6$
According to the question the ball attains its maximum height in $3{\text{ }}\sec $.
When a ball attains a maximum height then its final velocity becomes $0$.
Substituting the values in 1st Motion’s equation, $v = u + at$ where $v = $ final velocity, $u = $initial velocity, $a = $ acceleration and $t = $ time, we get
$0 = u - 12.6 \times 3$
$ \Rightarrow u = 37.8$
So, the initial velocity $u = 37.8{\text{ }}\dfrac{m}{s}$.
Alternative method:
We can also find $u$ from the derivative method. When the ball attains a maximum height then,
$\dfrac{{ds}}{{dt}} = 0$ from maximum property.
So, $u - 12.6 \times 3 = 0$
$ \Rightarrow u = 37.8$
Note:Students tend to make errors while solving these types of questions because they fail to apply the formula correctly, as well as to use the correct sign based on the direction of motion of the object. So, they should make sure to understand the problem's scenario clearly and then solve it without any calculation error.
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