Question

From the to p of tower156.8m high projectile is thrown up witha velocity of 39.2 making an angle 30 with horizontal direction find the distance from the foot of tower where it strickes the ground and the time taken by it

Answer 1

$$\begin{gathered} Initialvelocityinverticaldirection{u}_y=u\sin \theta \\ {u}_y=39.2\times \frac{1}{2}=19.6m/s \\ Initialvelocityinhorizontaldirection=u_x=u\cos \theta =39.2\times \frac{\sqrt{3}}{2}m/s \\ atthetopmostheightverticalcomponentofvelocityiszero. \\ {v}_y=0=u_y-gt \\ 0=19.6-9.8t \\ t=2s \\ dis\tan cetravelledin2s \\ s=u_{y}t-\frac{1}{2}g{t}^2 \\ s=19.6\times 2-0.5\times 9.8\times 2\times 2=19.6m \\ Netverticaldis\tan cetravelledindownwardmotion \\ d=19.6+156.8=176.4m \\ timetakentoreachtheground \\ d=ut+\frac{1}{2}g{t}^2 \\ whenbosystyartfallingu=0 \\ 176.4=0+0.5\times 9.8\times t^2 \\ t=6s \\ Horizontaldis\tan cetravelledin6+2=8s \\ R=u\cos \theta \times t \\ =39.2\times \frac{\sqrt{3}}{2}\times 8 \\ R=271.5m \end{gathered}$$