An object is thrown vertically upwards to a height of 10 m. Calculate its
(a) Velocity with which it was thrown upward.
(b) Time taken by the object to reach the highest point.
Answer
595.6k+ views
Hint: At the maximum height, the velocity of the object becomes zero. Use the kinematic equation which does not contain a time variable to determine the initial velocity of the object. To determine the time taken by the object to reach the highest point, use the kinematic equation which does not have the displacement variable.
Formula used:
Kinematic equation, \[{v^2} = {u^2} + 2as\]
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
\[v = u + at\]
Here, t is the time.
Complete step by step answer:
We have given the maximum vertical height attained by the object is \[h = 10\,m\].
(a)To calculate the initial velocity of the object with which it is thrown upward, we can use kinematic equation as follows,
\[{v^2} = {u^2} - 2gh\]
Here, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and h is the height attained by the object.
We know that at the maximum height, the velocity of the object becomes zero. Therefore, we can substitute 0 for v in the above equation.
\[{0^2} = {u^2} - 2gh\]
\[ \Rightarrow {u^2} = 2gh\]
\[ \Rightarrow u = \sqrt {2gh} \]
Substituting \[9.8\,m/{s^2}\] for g and 10 m for h in the above equation, we get,
\[u = \sqrt {2\left( {9.8} \right)\left( {10} \right)} \]
\[ \Rightarrow u = 14\,m/s\]
Therefore, the initial velocity of the object is 14 m/s.
(b) To calculate the time taken by the object to reach the height h, we can use following kinematic equation,
\[v = u - gt\]
Since final velocity v is zero, the above equation becomes,
\[u = gt\]
\[ \Rightarrow t = \dfrac{u}{g}\]
Substituting 14 m/s for u and \[9.8\,m/{s^2}\] for g in the above equation, we get,
\[t = \dfrac{{14}}{{9.8}}\]
\[ \Rightarrow t = 1.43\,{\text{s}}\]
Therefore, the time taken by the object to reach the highest point is 1.43 seconds.
Note:
Students should always remember that for the vertically upward motion of the body, the sign of the acceleration due to gravity must be negative and that for the downward motion, the sign of the acceleration due to gravity must be positive. To solve these types of questions, students should remember all the kinematic equations.
Formula used:
Kinematic equation, \[{v^2} = {u^2} + 2as\]
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
\[v = u + at\]
Here, t is the time.
Complete step by step answer:
We have given the maximum vertical height attained by the object is \[h = 10\,m\].
(a)To calculate the initial velocity of the object with which it is thrown upward, we can use kinematic equation as follows,
\[{v^2} = {u^2} - 2gh\]
Here, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and h is the height attained by the object.
We know that at the maximum height, the velocity of the object becomes zero. Therefore, we can substitute 0 for v in the above equation.
\[{0^2} = {u^2} - 2gh\]
\[ \Rightarrow {u^2} = 2gh\]
\[ \Rightarrow u = \sqrt {2gh} \]
Substituting \[9.8\,m/{s^2}\] for g and 10 m for h in the above equation, we get,
\[u = \sqrt {2\left( {9.8} \right)\left( {10} \right)} \]
\[ \Rightarrow u = 14\,m/s\]
Therefore, the initial velocity of the object is 14 m/s.
(b) To calculate the time taken by the object to reach the height h, we can use following kinematic equation,
\[v = u - gt\]
Since final velocity v is zero, the above equation becomes,
\[u = gt\]
\[ \Rightarrow t = \dfrac{u}{g}\]
Substituting 14 m/s for u and \[9.8\,m/{s^2}\] for g in the above equation, we get,
\[t = \dfrac{{14}}{{9.8}}\]
\[ \Rightarrow t = 1.43\,{\text{s}}\]
Therefore, the time taken by the object to reach the highest point is 1.43 seconds.
Note:
Students should always remember that for the vertically upward motion of the body, the sign of the acceleration due to gravity must be negative and that for the downward motion, the sign of the acceleration due to gravity must be positive. To solve these types of questions, students should remember all the kinematic equations.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

