Answer
Verified
312.9k+ views
Hint: Before you should solve the question, you should draw the diagram from the given information. So that it will be easier for us to solve the problem. The three equations of motion can be used to calculate the distance, velocity, position, and acceleration of the given object. At heights, the final velocity of the object becomes zero.
Complete answer:
It is shown in the below diagram that the window is located \[15.00m\] above the ground
So, \[s=15.00m\]
The three equations of motion are-
\[v=u+at\]……….(1)
\[{{v}^{2}}={{u}^{2}}+2as\]…….(2)
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]……….(3)
Where,
\[u=initial\text{ }velocity\]
\[v=final\text{ }velocity\]
\[t=time\]
\[s=dis\tan ce\]
\[a=acceleration\]
In the above question, it is given that a ball is thrown vertically upwards with a velocity \[5.00m/{{s}^{2}}\].
So the final velocity of the ball when going upwards will be \[v=5.00m/{{s}^{2}}\].
(a) In this part we have to find that how much distance does the ball covered in going upwards so for that total height of the ball should find out, which is as follows
we know that according to the second equation of motion
\[{{v}^{2}}={{u}^{2}}+2as\] …………..Eq(4)
And acceleration will be \[a=-10m/{{s}^{2}}\]
It is given that
(Acceleration will be negative because the ball is moving against gravity)
It is also given that, \[s=15.0m\]
Putting all these values in Eq(4 ), we get
\[\begin{align}
& {{v}^{2}}={{u}^{2}}+2as \\
& \Rightarrow {{(5)}^{2}}={{u}^{2}}-2\times 10\times 15 \\
& \Rightarrow 25={{u}^{2}}-300 \\
& \Rightarrow 25+300={{u}^{2}} \\
\end{align}\]
\[\begin{align}
& \Rightarrow {{u}^{2}}=325 \\
& \Rightarrow u=\sqrt{325}m/s \\
\end{align}\]
We know that the maximum height is given by –
\[H=\dfrac{{{u}^{2}}}{2g}\]………..Eq(5)
Where,
\[\begin{align}
& H=maximum\text{ }height \\
& u=initial\text{ }velocity \\
& g=acceleration \\
\end{align}\]
So after putting the value of \[u=\sqrt{325}m/s\] in eq(5), we can find the maximum height, which is as follows
\[\begin{align}
& H=\dfrac{{{u}^{2}}}{2g} \\
& \Rightarrow H=\dfrac{325}{2\times 10} \\
\end{align}\]
\[\Rightarrow H=16.5m\]
So the maximum height of the ball thrown upwards will be \[H=16.5m\].
(b) In this part we have to find the time taken by the ball to go from the ground to its highest point
When the ball reaches the highest point then its final velocity will be zero.
So, from the first equation of motion ,
\[v=u+at\]
Where
\[\begin{align}
& v=0m/s \\
& u=\sqrt{325}m/s \\
& a=-10m/{{s}^{2}} \\
\end{align}\]
On putting all these values in the above equation, we get
\[\begin{align}
& \Rightarrow 0=\sqrt{325}-10\times t \\
& \Rightarrow 10t=\sqrt{325} \\
& \Rightarrow t=\dfrac{\sqrt{325}}{10} \\
& \Rightarrow t=\dfrac{5\sqrt{13}}{10} \\
\end{align}\]
\[\Rightarrow t=\dfrac{\sqrt{13}}{2}\sec \]
\[\Rightarrow t=\dfrac{3.6}{2}\sec \]
(since the value of \[\sqrt{13}=3.6\])
\[\Rightarrow t=1.8\operatorname{s}\]
So the time taken by the ball to reach from ground to the highest point is \[t=1.8\operatorname{s}\].
Note: Retardation is another name for negative acceleration. Retardation is the opposite of acceleration which happens when the object moves opposite to the direction of velocity. Scalar quantities are the quantities that have only magnitude and no direction and vector quantities are the quantities that have both magnitude and direction.
Complete answer:
It is shown in the below diagram that the window is located \[15.00m\] above the ground
So, \[s=15.00m\]
The three equations of motion are-
\[v=u+at\]……….(1)
\[{{v}^{2}}={{u}^{2}}+2as\]…….(2)
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]……….(3)
Where,
\[u=initial\text{ }velocity\]
\[v=final\text{ }velocity\]
\[t=time\]
\[s=dis\tan ce\]
\[a=acceleration\]
In the above question, it is given that a ball is thrown vertically upwards with a velocity \[5.00m/{{s}^{2}}\].
So the final velocity of the ball when going upwards will be \[v=5.00m/{{s}^{2}}\].
(a) In this part we have to find that how much distance does the ball covered in going upwards so for that total height of the ball should find out, which is as follows
we know that according to the second equation of motion
\[{{v}^{2}}={{u}^{2}}+2as\] …………..Eq(4)
And acceleration will be \[a=-10m/{{s}^{2}}\]
It is given that
(Acceleration will be negative because the ball is moving against gravity)
It is also given that, \[s=15.0m\]
Putting all these values in Eq(4 ), we get
\[\begin{align}
& {{v}^{2}}={{u}^{2}}+2as \\
& \Rightarrow {{(5)}^{2}}={{u}^{2}}-2\times 10\times 15 \\
& \Rightarrow 25={{u}^{2}}-300 \\
& \Rightarrow 25+300={{u}^{2}} \\
\end{align}\]
\[\begin{align}
& \Rightarrow {{u}^{2}}=325 \\
& \Rightarrow u=\sqrt{325}m/s \\
\end{align}\]
We know that the maximum height is given by –
\[H=\dfrac{{{u}^{2}}}{2g}\]………..Eq(5)
Where,
\[\begin{align}
& H=maximum\text{ }height \\
& u=initial\text{ }velocity \\
& g=acceleration \\
\end{align}\]
So after putting the value of \[u=\sqrt{325}m/s\] in eq(5), we can find the maximum height, which is as follows
\[\begin{align}
& H=\dfrac{{{u}^{2}}}{2g} \\
& \Rightarrow H=\dfrac{325}{2\times 10} \\
\end{align}\]
\[\Rightarrow H=16.5m\]
So the maximum height of the ball thrown upwards will be \[H=16.5m\].
(b) In this part we have to find the time taken by the ball to go from the ground to its highest point
When the ball reaches the highest point then its final velocity will be zero.
So, from the first equation of motion ,
\[v=u+at\]
Where
\[\begin{align}
& v=0m/s \\
& u=\sqrt{325}m/s \\
& a=-10m/{{s}^{2}} \\
\end{align}\]
On putting all these values in the above equation, we get
\[\begin{align}
& \Rightarrow 0=\sqrt{325}-10\times t \\
& \Rightarrow 10t=\sqrt{325} \\
& \Rightarrow t=\dfrac{\sqrt{325}}{10} \\
& \Rightarrow t=\dfrac{5\sqrt{13}}{10} \\
\end{align}\]
\[\Rightarrow t=\dfrac{\sqrt{13}}{2}\sec \]
\[\Rightarrow t=\dfrac{3.6}{2}\sec \]
(since the value of \[\sqrt{13}=3.6\])
\[\Rightarrow t=1.8\operatorname{s}\]
So the time taken by the ball to reach from ground to the highest point is \[t=1.8\operatorname{s}\].
Note: Retardation is another name for negative acceleration. Retardation is the opposite of acceleration which happens when the object moves opposite to the direction of velocity. Scalar quantities are the quantities that have only magnitude and no direction and vector quantities are the quantities that have both magnitude and direction.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE