
A boy throws a ball in air in such a manner that when the ball is at its maximum height he throws another ball. If the balls are thrown with the time difference $1$ second, the maximum height attained by each ball is
(A) $9.8\,m$
(B) $19.6\,m$
(C) $4.9\,m$
(D) $2.45\,m$
Answer
222.6k+ views
Hint Use the formula of the acceleration given below and rearrange and substitute known values to find the initial velocity of the ball. Substitute this answer in the formula of the maximum height along with the acceleration due to gravity to find the maximum height.
Useful formula:
(1) The formula of the acceleration is given by
$a = \dfrac{u}{t}$
Where $a$ is the acceleration of the ball when it reaches maximum height, $u$ is the initial velocity of the ball at which it is thrown and $t$ is the time taken to reach the maximum height.
(2) The formula of the maximum height is given by
$h = \dfrac{{{u^2}}}{{2g}}$
Where $h$ is the maximum height of the ball, $g$ is the acceleration due to gravity.
Complete step by step answer
It is given that the
The time difference between the two balls, $t = 1\,s$
Since the second ball is thrown, when the first ball is at the maximum height, the difference in the time between the two balls must be equal to the time taken by the first ball to reach its maximum height. Let us use the formula of the acceleration, we get
$a = \dfrac{u}{t}$
Rearranging the equation and substituting the gravitation as the acceleration, since the balls are thrown against the gravity.
$t = \dfrac{u}{g} = \dfrac{u}{{9.8}}$
$1 = \dfrac{u}{{9.8}}$
By simplification of the above equation, we get
$u = 9.8\,m{s^{ - 1}}$
Let us use the formula of the maximum height.
$h = \dfrac{{{u^2}}}{{2g}}$
Substituting all the values known,
$h = \dfrac{{{{9.8}^2}}}{{2 \times 9.8}}$
By performing various arithmetic operations, we get
$h = 4.9\,m$
Hence the maximum height of the ball is obtained as $4.9\,m$ .
Thus the option (C) is correct.
Note The two balls will have an upward motion against the gravitational force until the force exerted on it greater than or equals the gravitational force. After that the force and the velocity of the balls becomes zero and return back to the earth with the acceleration due to the gravity.
Useful formula:
(1) The formula of the acceleration is given by
$a = \dfrac{u}{t}$
Where $a$ is the acceleration of the ball when it reaches maximum height, $u$ is the initial velocity of the ball at which it is thrown and $t$ is the time taken to reach the maximum height.
(2) The formula of the maximum height is given by
$h = \dfrac{{{u^2}}}{{2g}}$
Where $h$ is the maximum height of the ball, $g$ is the acceleration due to gravity.
Complete step by step answer
It is given that the
The time difference between the two balls, $t = 1\,s$
Since the second ball is thrown, when the first ball is at the maximum height, the difference in the time between the two balls must be equal to the time taken by the first ball to reach its maximum height. Let us use the formula of the acceleration, we get
$a = \dfrac{u}{t}$
Rearranging the equation and substituting the gravitation as the acceleration, since the balls are thrown against the gravity.
$t = \dfrac{u}{g} = \dfrac{u}{{9.8}}$
$1 = \dfrac{u}{{9.8}}$
By simplification of the above equation, we get
$u = 9.8\,m{s^{ - 1}}$
Let us use the formula of the maximum height.
$h = \dfrac{{{u^2}}}{{2g}}$
Substituting all the values known,
$h = \dfrac{{{{9.8}^2}}}{{2 \times 9.8}}$
By performing various arithmetic operations, we get
$h = 4.9\,m$
Hence the maximum height of the ball is obtained as $4.9\,m$ .
Thus the option (C) is correct.
Note The two balls will have an upward motion against the gravitational force until the force exerted on it greater than or equals the gravitational force. After that the force and the velocity of the balls becomes zero and return back to the earth with the acceleration due to the gravity.
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