So I'm doing my physics homework and I need some help on finding some answers. It's the last three questions on my sheet and I can't seem to get my head around it.
A ball is projected at $20\ m/s$ from the top of a $150\ m$ cliff. I need to calculate the Time Taken to hit the water at the bottom. Distance Travelled Horizontally and the Final Velocity. - This is an extension task I've never seen before and I don't want to give this up. How do I go about working this out?
HINTS: Using second equation of motion $$h=ut+\frac{1}{2}gt^2$$ setting $h=150\ m$, $u=20\ m/s$ & $g=10\ m/s^2$ we get $$t^2+4t-30=0$$ Find the time taken $t>0$ then $$\text{horizontal distance}=u_xt=20t$$ vertical component of velocity $v_y=u_yt+gt=10t$ The horizontal velocity $u_x=20\ m/s$ remains constant i.e. not affected by the gravity.
Hence the final velocity $$=\sqrt{(u_x)^2+(v_y)^2}=\sqrt{20^2+(10t)^2}$$