Question

A ball is thrown vertically upwards with a velocity of 6 ms-2. Calculate the maximum height to which it rises. (take acceleration due to gravity = 10 ms-2)

Answer 1

Answer:

the process by which green plants and some other organisms use sunlight to synthesize nutrients from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a by-product.

Answer 2

Answer:

The maximum height to which the ball rises is 1.8 m.

Step-by-step explanation:

${\underline{\underline{\bf{Given:}}}}$

• Final velocity of the ball = 6 m/s
• Initial velocity of the ball = 0 m/s
• Acceleration due to gravity = 10 m/s²

${\underline{\underline{\bf{To\:find:}}}}$

• The maximum height to which the ball rises.

${\underline{\underline{\bf{Concept:}}}}$

The 3 equations of motion:

‎ ‎ ‎ ‎ ‎ ‎ ‎$\sf{1)\:{\boxed{\sf{v=u+at}}}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎$\sf{2)\:{\boxed{\sf{s=ut+\dfrac{1}{2}at^2}}}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎$\sf{3)\:{\boxed{\sf{v^2-u^2=2as}}}}$

${\underline{\underline{\bf{Solution:}}}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎As we are given with values of initial velocity, final velocity and acceleration due to gravity, we can find the maximum height i.e, displacement by using the third equation of motion.

$\mapsto{\bf{\purple{v^2-u^2=2as}}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:v=Final\: velocity}$

$\:\:\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:u=Initial\: velocity}$

$\:\:\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:a=Acceleration}$

$\:\:\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:s=Displacement}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Here, the ball is thrown upwards which acts against the gravity. So, the value of acceleration due to gravity will be changed to negative value i.e, 10 m/s² changes to -10 m/s².

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Now, by substituting the given values in the equation:

$\:$

$\rightarrowtail\sf{(0)^2-(6)^2=2\times (-10)\times s}$

$\:$

And solving,

$\:\:\:\::\implies{\sf{0-36=2\times (-10)\times s}}$

$\:\:\:\:\:\:\:\::\implies{\sf{(-36)=2\times (-10)\times s}}$

$\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{(-36)=(-20)\times s}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{\dfrac{(\cancel{-}\:36)}{(\cancel{-}\:20)}=s}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{\dfrac{\cancel{36}}{\cancel{20}}=s}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\boxed{\frak{\red{1.8\:m=s}}}}$

$\:$

• Hence, the maximum height to which the ball rises is 1.8 m.

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