Rocket is fired from ground at a building 20,000 ft away. The initial velocity is $1,000 ft/s$ at an angle of $20^\circ$. How long will it take for the projectile to hit the building? I resolved the initial velocity triangle into the appropriate right triangle I use the equation $\Delta y=v_(iy)t + 1/2 at^2$. I get $t=21.375s$. Book gets $21.3$ so I assume that my answer is correct.
Then the question asks what will be the impact velocity. This is a math class. But in physics I was taught that not counting air friction, there would be a conservation of energy and the final velocity would be the initial velocity. So, I would think the answer would be $1000 f/s.$ The calculus book answer is $999.17$ at $160.13^\circ$.
Totally at a loss here as to how they get these numbers. I'm grasping at straws here, but I notice that the x direction of travel in 21.375 seconds is MORE than the 20,000 feet of the building. So, is the fact that the rocket hits the building before the 21.375 seconds factor into this? I've never encountered a problem like this.
The time of flight can be obtained by $$T=\frac{20000}{1000\cos 20^\circ}=21.284\text{ s}$$
The vertical velocity during impact is $$v_y(T)=v_y(0)-gT$$ $$=1000\sin 20^\circ -g\frac{20000}{1000\cos 20^\circ}$$
Taking $g=32\text{ ft/s}^2$, you have $$v_y(T)=-339.05363\text{ ft/s}$$
Of course, $v_x(T)=1000\cos 20^\circ=939.69262$
So
$$v(T)=\sqrt{939.69262^2+339.05363^2}=998.99\text{ ft/s}$$