Question

An object is thrown vertically upwards and rises to a height of 20 m. calculate i.) the velocity with which the object was thrown upwards,ii.) the time taken by the object to reach the highest point (g= 10 m/s^-2)for class 9th chapter gravitation

Answer 1

Given :

▪ An object is thrown vertically upward.

▪ Max. height reached = 20m

▪ Acc. due to gravity = 10m/s²

To Find :

▪ Initial velocity of ball with which the object was thrown upward.

▪ Time taken by ball to reach at the max. height.

ConcepT :

↗ Here, acceleration due to gravity has said to be constant, we can easily apply equation of kinematics to solve this type of questions.

↗ For a body falling freely under the action of gravity, g is taken positive.

↗ For a body thrown vertically upward, g is taken negative.

[At max. height v = 0]

✴ Max. height reached :

$\longrightarrow\sf\:v^2-u^2=2gs\\ \\ \longrightarrow\sf\:(0)^2-u^2=2(-g)H\\ \\ \longrightarrow\sf\:u^2=2gH\\ \\ \longrightarrow\underline{\boxed{\bf{\red{H=\dfrac{u^2}{2g}}}}}\:\gray{\bigstar}$

✴ Time of ascent :

$\Rightarrow\sf\:v=u+at\\ \\ \Rightarrow\sf\:0=u-gt\\ \\ \Rightarrow\sf\:u=gt\\ \\ \Rightarrow\underline{\boxed{\bf{\blue{t=\dfrac{u}{g}}}}}\:\gray{\bigstar}$

CalculaTion :

☣ Initial velocity :

$\mapsto\sf\:H=\dfrac{u^2}{2g}\\ \\ \mapsto\sf\:u=\sqrt{2gH}\\ \\ \mapsto\sf\:u=\sqrt{2\times 10\times 20}=\sqrt{400}\\ \\ \mapsto\underline{\boxed{\bf{\green{u=20\:ms^{-1}}}}}\:\orange{\bigstar}$

❇ Time of ascent :

$\dashrightarrow\sf\:t=\dfrac{u}{g}\\ \\ \dashrightarrow\sf\:t=\dfrac{20}{10}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{t=2s}}}}\:\orange{\bigstar}$

Answer 2

Given that ,

" An object is thrown vertically upwards and rises to a height of 20 m "

Here ,

Initial velocity , u = ? m/s

Final velocity , v = 0 m/s [ Since velocity at max. height is " 0 " ]

Acceleration , a = g = - 10 m/s² [ Since against gravity ]

Distance , s = 20 m

We need to find ,

___________________

( i ) The velocity with which the object was thrown upwards

Use 3rd equation of motion

⇒ v² - u² = 2as

⇒ 0² - u² = 2 ( - 10 ) ( 20 )

⇒ u² = 400

⇒ u = 20 m/s

_____________________

( ii )  The time taken by the object to reach the highest point

Use 1 st equation of motion

⇒ v = u + at

⇒ 0 = 20 - ( 10 ) t

⇒ 10t = 20

⇒ t = 2 s

____________________