Question

A shot is fired at an angle of 30 degrees with the horizontal from the top of a tower 182.88m. The velocity of projection is 60.90m/s . Find where from the foot of the tower it strikes the ground

Answer 1

• Given

        a shot is fired at 30°

        the top of the tower S = 182.88m

        the velocity of the projection u=60.9m/sec

• From the figure

       $u_{y}$=u*sinθ = 60.9*sin30 = 30.45 m/sec

       $u_{x}$=u*cosθ =60.9*cos30 = 52.7 m/sec

• Time to return from top of tower

        t = $\frac{2u_{y} }{g}$ = $\frac{2*30.45}{9.8}$ = 6.2 sec

• $v_{y}$ =$\sqrt{ u_{y} ^{2} + 2*g*S}$ = $\sqrt{ (30.45)^{2} + 2*9.8*182.88}$ = 67.1
• Time to reach the ground

       t' = $\frac{S}{\frac{u+v}{2} }$ = $\frac{182.88*2}{67.1+30.45}$ =3.7 sec

• Total time to reach the ground

       T= t + t' = 6.2 + 3.7 = 9.9 sec

• Horizontal range = $u_{x}$*t = 52.7*9.9 =521.73 m