What is the maximum height of a ball when thrown with $10m{{s}^{-1}}$ upwards?
Answer
578.1k+ views
Hint: The ball has been thrown up with some speed. So, the force of gravity will act on the ball in the downward direction. This is a constant force due to Earth’s pull. Also, at its maximum height, the speed of the ball will be equal to zero. So, we can use the equation of “straight line motion” with constant acceleration to get our answer.
Complete answer:
Let us first assign some terms that we are going to use in our solution.
Let the initial speed of the ball before releasing be ‘u’. Then, the value of ‘u’ has been given to us as:
$\Rightarrow u=10m{{s}^{-1}}$
Let the final speed of the ball, that is, the speed of the ball at its maximum height be ‘v’. Then, the value of ‘v’ will be equal to:
$\Rightarrow v=0m{{s}^{-1}}$
Let the acceleration due to gravity be denoted by ‘g’. Then, the value of ‘g’ is known to us as:
$\Rightarrow g=10m{{s}^{-2}}$
Now, let the maximum height attained by the ball be ‘h’. Then, we can calculate this height with the help of following formula:
$\Rightarrow {{v}^{2}}={{u}^{2}}+2gh$
Putting the values of all the known terms from above equations and calculating for ‘g’, we get:
$\begin{align}
& \Rightarrow {{0}^{2}}={{10}^{2}}+2\times 10\times h \\
& \Rightarrow -100=20h \\
& \Rightarrow h=\dfrac{-100}{20} \\
& \therefore h=-5m \\
\end{align}$
Here, the negative sign in height means that the ball has travelled this distance opposite to the direction of gravity (which was taken as positive in our above equation).
Hence, the maximum height of a ball when thrown with $10m{{s}^{-1}}$ upwards comes out to be 5 meters.
Note:
Since, there is no mention of air resistance, we will always consider our motion in a frictionless environment. Also, the other method to solve this problem would have been the energy conservation method. By conserving energy at the time of release and at the time when the ball is at its maximum height, we would have got the same result as we did by applying the equation of motion with constant acceleration.
Complete answer:
Let us first assign some terms that we are going to use in our solution.
Let the initial speed of the ball before releasing be ‘u’. Then, the value of ‘u’ has been given to us as:
$\Rightarrow u=10m{{s}^{-1}}$
Let the final speed of the ball, that is, the speed of the ball at its maximum height be ‘v’. Then, the value of ‘v’ will be equal to:
$\Rightarrow v=0m{{s}^{-1}}$
Let the acceleration due to gravity be denoted by ‘g’. Then, the value of ‘g’ is known to us as:
$\Rightarrow g=10m{{s}^{-2}}$
Now, let the maximum height attained by the ball be ‘h’. Then, we can calculate this height with the help of following formula:
$\Rightarrow {{v}^{2}}={{u}^{2}}+2gh$
Putting the values of all the known terms from above equations and calculating for ‘g’, we get:
$\begin{align}
& \Rightarrow {{0}^{2}}={{10}^{2}}+2\times 10\times h \\
& \Rightarrow -100=20h \\
& \Rightarrow h=\dfrac{-100}{20} \\
& \therefore h=-5m \\
\end{align}$
Here, the negative sign in height means that the ball has travelled this distance opposite to the direction of gravity (which was taken as positive in our above equation).
Hence, the maximum height of a ball when thrown with $10m{{s}^{-1}}$ upwards comes out to be 5 meters.
Note:
Since, there is no mention of air resistance, we will always consider our motion in a frictionless environment. Also, the other method to solve this problem would have been the energy conservation method. By conserving energy at the time of release and at the time when the ball is at its maximum height, we would have got the same result as we did by applying the equation of motion with constant acceleration.
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