Answer to Question #113589 in Atomic and Nuclear Physics for God

Question

A cannon fire a ball off of a 20.0 m high cliff. What is the range of the cannon if fired at 100.0 m/s at a 30.0 degree angle?

Answer 1

here the total range will be R1+R2

we have to calculate R1+R2

first let's calculate R1

R1=$v^2\sin2\theta/2g=\frac{100\times100\times\sqrt3}{2\times10\times2}=250\sqrt3m$

and now let's calculate R2

the horizontal component of velocity=+30 cos30=15$\sqrt3$ m/s

vertical component of velocity =30 sin30=15 m/s

applying S=ut+0.5gt²

20=15t+5t²

on solving we get t=1 seconds (although it has two roots but the other one is negative and we know time cannot we negative )

R2=15$\sqrt3$ *1=15$\sqrt3$m

hence total range R1+R2 = 265$\sqrt3$ m