Physics - Kinematics Question with Solution | TestHub

$$u_y=\sqrt{2gH}=\sqrt{2(10)(100)}=20\sqrt{5}m/s$$

At the second pole the ball is descending at 45°, hence its downward speed equals ux.

The poles have equal heights, so their crossing times lie symmetrically about the apex.

$$100=\frac{2u_x^2}{g}$$

$$u_x=10\sqrt{5}m/s$$

$$\theta ={\tan }^{-1}\left(\frac{u_y}{u_x}\right)={\tan }^{-1}2\text{(C)}$$

$$T_{\mathrm{flight}}=\frac{2u_y}{g}=4\sqrt{5}\mathrm{s}\text{(A)}$$

$$\Delta t_{\mathrm{poles}}=\frac{100}{u_x}=2\sqrt{5}\mathrm{s}\text{(B)}$$

The second pole is crossed Δt/2=√5 s after the apex.

$$H_{\mathrm{pole}}=100-\frac{1}{2}g(\sqrt{5}{)}^2$$

$$H_{\mathrm{pole}}=100-25=75\mathrm{m}\text{(D)}$$