Answer
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Hint: When a ball is thrown upwards under earth’s gravity, it faces deceleration due to the acceleration due to gravity and air resistance is also present here which decreases the speed of the ball and we can describe the motion of the ball using equations of motion.
Complete step-by-step answer:
We have a ball which is being thrown upwards with velocity 10 m/s, this means that our initial velocity is u = 10 m/s.
Now as the ball goes up, its speed decreases due to two reasons:
Gravitational pull of earth on the ball – acceleration due to gravity tends to decrease the speed during upward motion and increase the speed during downward motion.
The resistance of the air – air acts as a viscous fluid which tries to decelerate the ball
Let the acceleration due to gravity be g. If the resistive force of air is given as R = ma where a is deceleration due to air resistance.
We can use the following equation of motion to describe the motion of a ball.
${{\text{v}}^2} - {u^2} = 2aS$
For upward motion, u = 10 m/s, v = 0 (as at the top ball comes to rest momentarily), S = h ( the height attained by the ball)
$0 = {u^2} - 2\left( {g + \dfrac{R}{m}} \right)h{\text{ }}$
where total deceleration is equal to the sum of those due to gravity and air resistance
$0 = {\left( {10} \right)^2} - 2\left( {g + \dfrac{R}{m}} \right)h{\text{ }}...{\text{(i) }}$
For downward motion, v = 8 m/s, u = 0
$
0 = {{\text{v}}^2} - 2\left( {g - \dfrac{R}{m}} \right)h{\text{ }} \\
\Rightarrow 0 = {\left( 8 \right)^2} - 2\left( {g - \dfrac{R}{m}} \right)h{\text{ }}...{\text{(ii) }} \\
$
Here g and R/m act in opposite directions
Adding the equations (i) and (ii), we get
$
100 + 64 - 4gh = 0 \\
\Rightarrow 4gh = 164 \\
\Rightarrow h = \dfrac{{164}}{{4g}} \\
$
Taking $g = 10m{s^{ - 2}}$, we get
$
h = \dfrac{{164}}{{40}} \\
\Rightarrow h = 4.1m \\
$
Hence, the correct answer is option D.
Note: If the body is accelerating then the acceleration is taken to be positive but if the force is decelerating a body then its value is taken to be negative in equations of motions the body under consideration.
Complete step-by-step answer:
We have a ball which is being thrown upwards with velocity 10 m/s, this means that our initial velocity is u = 10 m/s.
Now as the ball goes up, its speed decreases due to two reasons:
Gravitational pull of earth on the ball – acceleration due to gravity tends to decrease the speed during upward motion and increase the speed during downward motion.
The resistance of the air – air acts as a viscous fluid which tries to decelerate the ball
Let the acceleration due to gravity be g. If the resistive force of air is given as R = ma where a is deceleration due to air resistance.
We can use the following equation of motion to describe the motion of a ball.
${{\text{v}}^2} - {u^2} = 2aS$
For upward motion, u = 10 m/s, v = 0 (as at the top ball comes to rest momentarily), S = h ( the height attained by the ball)
$0 = {u^2} - 2\left( {g + \dfrac{R}{m}} \right)h{\text{ }}$
where total deceleration is equal to the sum of those due to gravity and air resistance
$0 = {\left( {10} \right)^2} - 2\left( {g + \dfrac{R}{m}} \right)h{\text{ }}...{\text{(i) }}$
For downward motion, v = 8 m/s, u = 0
$
0 = {{\text{v}}^2} - 2\left( {g - \dfrac{R}{m}} \right)h{\text{ }} \\
\Rightarrow 0 = {\left( 8 \right)^2} - 2\left( {g - \dfrac{R}{m}} \right)h{\text{ }}...{\text{(ii) }} \\
$
Here g and R/m act in opposite directions
Adding the equations (i) and (ii), we get
$
100 + 64 - 4gh = 0 \\
\Rightarrow 4gh = 164 \\
\Rightarrow h = \dfrac{{164}}{{4g}} \\
$
Taking $g = 10m{s^{ - 2}}$, we get
$
h = \dfrac{{164}}{{40}} \\
\Rightarrow h = 4.1m \\
$
Hence, the correct answer is option D.
Note: If the body is accelerating then the acceleration is taken to be positive but if the force is decelerating a body then its value is taken to be negative in equations of motions the body under consideration.
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