Question

A body projected vertically upward with a velocity u reaches a maximum height h . The velocity of projection to reach double the maximum height i.e., 2h is

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Initial\:velocity=\sqrt{2}u\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Maximum \: height = 2h \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Initial \: velocity =?$

• According to given question :

$\tt \circ \: Final \: velocity = 0 \: m/s \\ \\ \bold{as \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {u}^{2} + 2 \times a \times h \\ \\ \tt: \implies - {u}^{2} = 2h \times a \\ \\ \tt: \implies a = - \frac{ {u}^{2} }{2h}\:m/s^{2}$

$\tt \circ \: Acceleration = - \frac{{u}^{2} }{2h} \:m/s^{2} \\ \\ \bold{as \: we \: know \: that} \\ \tt: \implies {v}^{2} = { u_{1}}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = { u_{1}}^{2} + 2 \times ( - \frac{ {u}^{2} }{2h} ) \times 2h \\ \\ \tt: \implies - { u_{1}}^{2} = - 2 {u}^{2} \\ \\ \tt: \implies { u_{1}}^{2} =2 {u}^{2} \\ \\ \green{\tt: \implies { u_{1}} = \sqrt{2}u \: m/s } \\ \\ \green{\tt \therefore initial \: velocity \: to \: reach \: \: 2h \: \: height \: is \: \: \sqrt{2}u \: \: m/s}$

Answer 2

Hope this helps uh....!