Question

a stone is thrown vertically upwards such that it takes 2 sec to attain max height.find max height attained by the ball.(acceleration due to gravity=9.8m/)

Answer 1

Given

A stone is thrown vertically upwards such that it takes 2 sec to attain max height

To Find

Max height attained by the ball

Solution

Given ,

• Time , t = 2 s
• Final velocity , v = 0 m/s

∵ Goes to extreme position

• Acceleration , a = - 9.8 m/s²

Apply 1st equation of motion ,

$\bf \pink{\bigstar\ \; v=u+at}\\\\\to \rm 0=u+(-9.8)(2)\\\\\leadsto \bf \green{u=19.6\ m/s\ \; \bigstar}$

Apply 3rd equation of motion ,

$\bf \red{\bigstar\ \; v^2-u^2=2as}\\\\\to \rm (0)^2-(19.6)^2=2(-9.8)s\\\\\to \rm 0-(19.6)^2=-19.6s\\\\\to \rm -(19.6)^2=-19.6s\\\\\leadsto \bf \blue{s=19.6\ m\ \; \bigstar}$

So , Maximum height attained by the ball is 19.6 m

Answer 2

Answer:

Given

A stone is thrown vertically upwards such that it takes 2 sec to attain max height

To Find

Max height attained by the ball

Solution

Given ,

Time , t = 2 s

Final velocity , v = 0 m/s

∵ Goes to extreme position

Acceleration , a = - 9.8 m/s²

Apply 1st equation of motion ,

\begin{gathered}\bf \pink{\bigstar\ \; v=u+at}\\\\\to \rm 0=u+(-9.8)(2)\\\\\leadsto \bf \green{u=19.6\ m/s\ \; \bigstar}\end{gathered}

★ v=u+at

→0=u+(−9.8)(2)

⇝u=19.6 m/s ★

Apply 3rd equation of motion ,

\begin{gathered}\bf \red{\bigstar\ \; v^2-u^2=2as}\\\\\to \rm (0)^2-(19.6)^2=2(-9.8)s\\\\\to \rm 0-(19.6)^2=-19.6s\\\\\to \rm -(19.6)^2=-19.6s\\\\\leadsto \bf \blue{s=19.6\ m\ \; \bigstar}\end{gathered}

★ v

2

−u

2

=2as

→(0)

2

−(19.6)

2

=2(−9.8)s

→0−(19.6)

2

=−19.6s

→−(19.6)

2

=−19.6s

⇝s=19.6 m ★

So , Maximum height attained by the ball is 19.6 m