You visited us 1 times! Enjoying our articles? Unlock Full Access!

Pure Rolling

Maximum heigh...

Question

Maximum height reached by ball $H$ above ground is

\frac{9 h}{35}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{18 h}{35}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{18 h}{25}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{27 h}{125}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{27 h}{125}$
$H = \frac{v^{2} {sin}^{2} sin 37^{\circ}}{2 g}$
$= \frac{(6 g h / 5) (3 / 5)^{2}}{2 g} = \frac{27 h}{125}$

Suggest Corrections

Similar questions

h

t

h (t) = - 16 t^{2} + 46 t + 5

\frac{- 18 h^{4} k^{5}}{- 6 h^{2} k^{2}}

600 g

25 m

35 %

h

T

(m)

\frac{T}{3}

Join BYJU'S Learning Program