Question

A particle is projected downwards from point 'A' with initial velocity $5m/s$ at an angle $\theta ={37}^{o}C$ with the vertical. A strong horizontal wind gives the particle a constant horizontal acceleration of $6m/s^2$ in the x - direction. If the particle strikes the ground directly under its released position, the height 'h' of point A (in $m$) is (answer upto two decimal places). The downward acceleration is taken as $g=10m/s^2$ , take (sin37°=3/5, cos37°=4/5)

Answer 1

$\begin{array}{c}\text{For A to B}\\ s_x=u_{x}t+\frac{1}{2}{a}_x^2\\ 0=-\text{using}t+\frac{1}{2}a{t}^2\\ t=\frac{2u\sin \theta }{a}=\frac{2\left(5\right)\sin 37°}{6}=1\text{sec.}\\ s_y=h=AB=u_{y}t+\frac{1}{2}{a}_y{t}^2\\ =u\cos \theta t+\frac{1}{2}g{t}^2\\ =\left(5\right)\left(\frac{4}{5}\right)\left(1\right)+\frac{1}{2}(10\left(1\right))\\ =9m\end{array}$