Question

a kangaroo can jump 2.5 m vertically. what must be the initial velocity of the kangaroo?​

Answer 1

Answer: 7 m/sec

Using third equation of motion,

v^2 = u^2 + 2as

v is equal to zero since velocity at highest point = 0

0 = u^2 + 2 ( - 9.8 ) × 2.5

0 = u^2 - 49

u^2 = 49

u = √49

u = 7 m/sec

Answer 2

$\sf \huge\underline{solution} \implies$

$\implies{ \sf{given}} \rightarrow$

$\implies {\sf{acceleration \: (a) = 9.8 \: m/s {}^{2} }}$

$\implies{\sf{distance \: (s) = 2.5 \: m}}$

$\implies{\sf{final \: velocity \: (v) = 0}}$

$\implies{ \sf{to \: find}} \rightarrow$

$\implies{\sf{initial \: velocity \: (u)}}$

$\implies{ \sf{formula}} \rightarrow$

$\implies{ \sf{v {}^{2} = u {}^{2} + 2as }}$

$\implies{ \sf{calculation}} \rightarrow$

$\implies{ \sf{from \: formula}} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: (0) {}^{2} = u {}^{2} + 2 \times ( - 9.8) \: (2.5)} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: here \: negative \: sign \: indicates \: that} \\ \sf{\: \: \: \: \: \: \: \: \: \: \: \: acceleration \: and \: velocity \: are \: in \: the \: oposite \: direction .}$

$\implies{ \sf{0 = u {}^{2} - 49}}$

$\implies{ \sf{u {}^{2} = 49 }}$

$\implies{ \sf{u = \fbox{7 \: m/s}}}$

$\implies{ \sf \therefore \: {the \: initial \: velocity \: of \: kangaroo \: must \: be \: 7 \: m/s.}}$