
A wooden block is dropped from the top of a cliff $100m$ high and simultaneously a bullet of mass $10g$ is fired from the foot of the cliff upwards with a velocity of $100m{s^{ - 1}}$ The bullet and wooden block will meet after a time.
(A) $10s$
(B) $0.5s$
(C) $1s$
(D) $7s$
Answer
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Hint:In order to solve this question, we will use the concept that, at the point of meeting of bullet and wooden block the sum of distance covered by both block and bullet will be equal to height of cliff and we will use equations of motion and will solve for time when they met during the journey.
Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$ where,
S is the distance covered by the body during time ‘t’ with initial velocity ‘u’ and acceleration of the body is ‘a’.
Complete answer:
According to the question, we have given that total height of the cliff is $h = 100m$ now, for the wooden block which is dropped under the effect of gravity so, for wooden block we have
$u = 0$ initial velocity is zero
$a = g = 10m{s^{ - 2}}$ acceleration due to gravity
and let ‘S’ be the distance covered by the wooden block in time t, so using the formula $S = ut + \dfrac{1}{2}a{t^2}$ and putting the required values, we get
$
S = \dfrac{1}{2}(10){t^2} \\
S = 5{t^2} \to (i) \\
$
Now, when bullet is fired vertically upward with velocity $u' = 100m{s^{ - 1}}$
and acceleration due to gravity is in opposite direction of motion, so $a = g = - 10m{s^{ - 2}}$ let the bullet covers a distance of S’ in same time ‘t’ in order to meet the wooden block so again using the formula $S = ut + \dfrac{1}{2}a{t^2}$ and putting the values we get,
$
S' = 100t - \dfrac{1}{2}(10){t^2} \\
S' = 100t - 5{t^2} \to (ii) \\
$
Now, we know that at moment of meeting between block and bullet, sum of their distances is equal to height of cliff so,
$h = S + S'$ ,on putting the values of h, S, and S’ we get
$
100 = 5{t^2} + 100t - 5{t^2} \\
100 = 100t \\
\Rightarrow t = 1s \\
$
So, Wooden block and the bullet will meet after a time of $1s$
Hence, the correct option is Option (C).
Note:It should be noted that, while solving such questions don’t mistake sign conventions used for acceleration due to gravity, if motion is in the direction of gravity then acceleration is taken as positive while it’s negative during motion in opposite direction to that of gravity.
Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$ where,
S is the distance covered by the body during time ‘t’ with initial velocity ‘u’ and acceleration of the body is ‘a’.
Complete answer:
According to the question, we have given that total height of the cliff is $h = 100m$ now, for the wooden block which is dropped under the effect of gravity so, for wooden block we have
$u = 0$ initial velocity is zero
$a = g = 10m{s^{ - 2}}$ acceleration due to gravity
and let ‘S’ be the distance covered by the wooden block in time t, so using the formula $S = ut + \dfrac{1}{2}a{t^2}$ and putting the required values, we get
$
S = \dfrac{1}{2}(10){t^2} \\
S = 5{t^2} \to (i) \\
$
Now, when bullet is fired vertically upward with velocity $u' = 100m{s^{ - 1}}$
and acceleration due to gravity is in opposite direction of motion, so $a = g = - 10m{s^{ - 2}}$ let the bullet covers a distance of S’ in same time ‘t’ in order to meet the wooden block so again using the formula $S = ut + \dfrac{1}{2}a{t^2}$ and putting the values we get,
$
S' = 100t - \dfrac{1}{2}(10){t^2} \\
S' = 100t - 5{t^2} \to (ii) \\
$
Now, we know that at moment of meeting between block and bullet, sum of their distances is equal to height of cliff so,
$h = S + S'$ ,on putting the values of h, S, and S’ we get
$
100 = 5{t^2} + 100t - 5{t^2} \\
100 = 100t \\
\Rightarrow t = 1s \\
$
So, Wooden block and the bullet will meet after a time of $1s$
Hence, the correct option is Option (C).
Note:It should be noted that, while solving such questions don’t mistake sign conventions used for acceleration due to gravity, if motion is in the direction of gravity then acceleration is taken as positive while it’s negative during motion in opposite direction to that of gravity.
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