Question

A projectile is projected from a point 0 on the ground
with an initial velocity u at an elevation angle e from the
horizontal direction as shown in the figure. It just crosses
two walls A and B of same height h situated symmetrically at
times ty=2 s and t2=6 s respectively. The horizontal distance
between the two walls is d = 120 m. You may presume
g = 10 m s-2​

Answer 1

Given:

$\text { The projectile is at a height } \mathrm{h} \text { at } \mathrm{t}=2 \text { and } \mathrm{t}=6 \text { seconds . }$

$\mathrm{h}=\mathrm{u} \sin \theta \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$

$\text { at } \mathrm{t}=2 \text {, }$

$\mathrm{h}=2 \mathrm{u} \sin \theta-2 \mathrm{~g}$

$\text { at } t=6 \text {, }$

$\mathrm{h}=6 \mathrm{u} \sin \theta-18 \mathrm{~g}$

$\therefore 4 \mathrm{u} \sin \theta=16 \mathrm{~g}$

$\mathrm{u} \sin \theta=4 \mathrm{~g}=40 \ldots(\mathrm{i})$

$\text { As the separation between the } 2 \text { points is } 120 \mathrm{~m} \text {, }$

$\begin{array}{l}\mathrm{u} \cos \theta \times \Delta \mathrm{t}=120 \\4 \mathrm{u} \cos \theta=120 \\\mathrm{u} \cos \theta=30 \ldots \text { (ii) }\end{array}$

$\text { Dividing equation (i) by (iii) }$

$\tan \theta=\frac{4}{3}$

Hence the answer is  $\tan \theta=\frac{4}{3}$