Question

A ball is thrown from the rod of a building of height $44m$ with speed $v_0$ at an angle $\theta$ below the horizontal. It lands $2$ seconds later at a point $30m$ from the base of the building, then the value of $\tan \theta$ is: $(g=10m/s^2)$

• A. $\frac{4}{5}$
• B. $\frac{3}{5}$
• C. $\frac{5}{4}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{4}{5}$

$\left(A\right)$

using : $v=u+at$ & $s=ut+\frac{a{t}^2}{2}$

$s_x=30,s_y=44,u_x=v_{o}cos\theta ,u_y=v_{o}sin\theta ,t=2$

$s_x=u_x\times t$

$30=v_{o}cos\theta \times 2$

$v_{o}cos\theta =15\dots \left(1\right)$

$s_y=u_{y}t+\frac{g{t}^2}{2}$

$44=v_{o}sin\theta \times 2+2g$

$44-20=2v_{o}sin\theta$

$v_{o}sin\theta =12\dots \left(2\right)$

Dividing $\left(2\right)$by $\left(1\right)$

$tan\theta =\frac{12}{15}=\frac{4}{5}$