Question

a bomber is flying upwards at an angle 53 degree with vertical ,releases a bomb at 800m altitude. the bomb strikes ground 20 sec after its release

find

velocity of bomber at the time of release of bomb

Answer 1

Let the component of speed in y direction be $u_y$

Given,

$g = -10 m/sec^{2} \\ \\ h = 800m \\ \\ t = 20 sec \\ \\Using \ second \ equation \ of \ motion \\ \\ h = u_y + \frac{1}{2}gt^{2} \\ \\ -800 = u_y*20 - \ \frac{1}{2}* 10*20^{2} \\ \\ -800 = 20u_y - 2000 \\ \\ 1200 = 20u_y \\ \\ u_y = 60 \ m/sec \\ \\ \text{ We know that , if velocity of object is u } , \\ \\ u_y = u \cos53 \\ \\ 60 = 0.6*u \\ \\ u = 600/6 = 100 \\ \\ u = 100 \ m/sec$

Answer 2

Given data states that the bomber flying at an angle 53° with bomb released at an altitude of 800 m denoted by h and strikes ground in time t = 20 sec . therefore by equation of motion,

                 $h = ut + \frac {1} {2} a t^2 \\\Rightarrow h = u y t + \frac {1} {2} g t^2 \\\Rightarrow - 800 = uy \times 20 + \frac {1} {2} ( - 10 ) \times ( 20 )^2\\ \Rightarrow - 800 = 20 u y - 2000 \\\Rightarrow20 u y = 1200\\ \Rightarrowuy\quad =\quad ucos35^{\circ}.$

Therefore u y is the vertical component,  

                $uy\quad =\quad ucos35^{\circ}\\\Rightarrow 60 = u cos 53 degree\\\Rightarrow u\quad =\quad 100\frac { m }{ s }$