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Answer 1

Answer:

Question :-

• A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo ?

Given :-

• A kangaroo is capable of jumping to a height of 2.62 m.

To Find :-

• What is the takeoff speed of the kangaroo.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2gh}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity (9.8 m/s²)
• h = Height

Solution :-

Given :

$\bigstar\: \: \bf{Final\: Velocity\: (v) =\: 0\: m/s}$

$\bigstar\: \: \bf{Acceleration\: due\: to\: gravity\: (g) =\: -\: 9.8\: m/s^2}$

[ Note : Acceleration due to gravity i.e, g = - 9.8 m/s², we take negetive sign because of downward direction. ]

$\bigstar\: \bf{Height\: (h) =\: 2.62\: m}$

❒ According to the question by using the formula we get,

$\implies \sf (0)^2 =\: u^2 + 2 \times (- 9.8) \times 2.62$

$\implies \sf 0 \times 0 =\: u^2 + 2 \times \bigg(- \dfrac{98}{10}\bigg) \times \bigg(\dfrac{262}{100}\bigg)$

$\longrightarrow \sf 0 =\: u^2 + 2 \times \bigg(- \dfrac{25676}{1000}\bigg)$

$\longrightarrow \sf 0 - u^2 =\: \bigg(- \dfrac{51352}{1000}\bigg)$

$\longrightarrow \sf {\cancel{-}} u^2 =\: {\cancel{-}} 51.352$

$\longrightarrow \sf u^2 =\: 51.352$

$\longrightarrow \sf u =\: \sqrt{51.352}$

$\longrightarrow \sf \bold{\red{u =\: 7.17\: m/s}}$

${\small{\bold{\purple{\underline{\therefore\: The\: takeoff\: speed\: of\: the\: kangaroo\: is\: 7.17\: m/s\: .}}}}}$

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$\leadsto\large\tt{\green{EXTRA\: INFORMATION\: :-}}$

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Second Equation Of Motion Formula :

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

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken
• s = Distance Covered or Displacement

Answer 2

Siso here is your answer