Question

Answer it please.....​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• acceleration " a " = 19.6 m/sec²
• Time " t " = 5 sec
• Initial velocity " u " = 0 m/sec

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Maximum height of the rocket from earth's surface

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

We know,

Newton's second low of motion

$\small\boxed{\sf{\:s\:=\:ut+\dfrac{1}{2}\times a.t^2}}$

where,

• s = Distance

keep all values,

$\mapsto\sf{\:s\:=\:0\times5+\dfrac{1}{2}\times (19.6)\times (5)^2} \\ \\ \\ \\ \mapsto\sf{\:s\:=\:0+9.8\times 25} \\ \\ \\ \mapsto\sf{\:s\:=\:245\:meter}$

$\Large{\underline{\mathfrak{\bf{\Hence}}}}$

$\mapsto\textsf{\:Maximum height of the} \\ \textsf{\:rocket from earth's surface will be \:=\:245\:meter}$

So, Our answer will be option number (a)

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$