
A stone is thrown in vertically upward direction with a velocity of \[5m{{s}^{-1}}\]. If the acceleration of the stone during its motion is \[10m{{s}^{-2}}\] in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
A. Height = 1.25 m and time 0.5 s
B. Height = 2.5 m and time 0.5 s
C. Height = 2.5 m and time 0.05 s
D. Height = 1.25 m and time 0.05 s
Answer
522.7k+ views
Hint: Recall the equations of motion and apply the suitable equations consecutively for finding height attained by the stone and time taken. Equations of motion are as follows:
\[v=u+at\]
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
\[{{v}^{2}}={{u}^{2}}+2as\]
Complete step by step answer:
Given that,
Initial velocity, \[u=5m{{s}^{-1}}\]
Final velocity, \[v=0\]
Here the initial velocity is in upward direction and acceleration in downward so the acceleration will come with negative sign, i.e., retarded motion.
\[a=-10m{{s}^{-2}}\]
Now we have to find out the height attained by stone \[\left( h \right)\] and time taken \[\left( t \right)\]
For calculating the height attained on applying 3rd equation of motion,
\[\Rightarrow {{v}^{2}}={{u}^{2}}+2as\]
\[\Rightarrow 0={{\left( 5 \right)}^{2}}+2\times \left( -10 \right)\times h\]
\[\Rightarrow 20h=25\]
\[\Rightarrow h=1.25m\]
Now for calculating the time period, on applying 1st equation of motion,
\[\Rightarrow v=u+at\]
\[\Rightarrow 0=5+\left( -10 \right)t\]
\[\Rightarrow 10t=5\]
\[\Rightarrow t=\dfrac{5}{10}\]
\[\Rightarrow t=0.5s\]
Hence, the correct option is A, i.e., 1.25 m and time 0.5 s
Additional Information:
(1). Students should proceed in the following manner for solving problems related to the equation of motion:
(2). Read the question carefully to identify the given quantities and note them.
Identify the equation to use and note them.
(3). Ensure that all the values are in the same system of measurement and put them in identified equations.
(4). Calculate the answer carefully and check the final units.
Note: Students should memorize the three equations of motion and understand the physical significance of these equations. Students need to know how to apply the suitable equation of motion on the basis of given data. Students should ensure that all the quantities are in the same system of measurement i.e., S.I., M.K.S., C.G.S. etc.
\[v=u+at\]
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
\[{{v}^{2}}={{u}^{2}}+2as\]
Complete step by step answer:
Given that,
Initial velocity, \[u=5m{{s}^{-1}}\]
Final velocity, \[v=0\]
Here the initial velocity is in upward direction and acceleration in downward so the acceleration will come with negative sign, i.e., retarded motion.
\[a=-10m{{s}^{-2}}\]
Now we have to find out the height attained by stone \[\left( h \right)\] and time taken \[\left( t \right)\]
For calculating the height attained on applying 3rd equation of motion,
\[\Rightarrow {{v}^{2}}={{u}^{2}}+2as\]
\[\Rightarrow 0={{\left( 5 \right)}^{2}}+2\times \left( -10 \right)\times h\]
\[\Rightarrow 20h=25\]
\[\Rightarrow h=1.25m\]
Now for calculating the time period, on applying 1st equation of motion,
\[\Rightarrow v=u+at\]
\[\Rightarrow 0=5+\left( -10 \right)t\]
\[\Rightarrow 10t=5\]
\[\Rightarrow t=\dfrac{5}{10}\]
\[\Rightarrow t=0.5s\]
Hence, the correct option is A, i.e., 1.25 m and time 0.5 s
Additional Information:
(1). Students should proceed in the following manner for solving problems related to the equation of motion:
(2). Read the question carefully to identify the given quantities and note them.
Identify the equation to use and note them.
(3). Ensure that all the values are in the same system of measurement and put them in identified equations.
(4). Calculate the answer carefully and check the final units.
Note: Students should memorize the three equations of motion and understand the physical significance of these equations. Students need to know how to apply the suitable equation of motion on the basis of given data. Students should ensure that all the quantities are in the same system of measurement i.e., S.I., M.K.S., C.G.S. etc.
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