Answer to Question #226693 in Civil and Environmental Engineering for Rexie

Question

What is the horizontal range of projectile if the object is fired from a building 400m high and has an initial velocity of 350m/s? The angle of elevation of the projectile is 250?

Answer 1

$θ=25° \\ R= u \cos θ×T\\ T=total \space time \space of \space flight\\ T=\frac{2u sin θ}{R}\\ R=u \cos θ× \frac{2u sin θ}{g} \\ R =\frac{u^2 sin 2θ}{g}\\ R =\frac{350^2 sin 50}{9.8}=9575.55 m$

This R is the horizontal range of the projectile when it is at the level of  400 m above the ground, 

At this point  the velocities are again u and making an angle 25° with the horizontal but pointing downwards

$400=350 sin 25×t+\frac{1}{2}×9.8×t^2\\ t=2.497 s$

During this time interval, the extra range covered will be

$r=u cos θ×t \\ =350 cos 25×2.497 \\ =792.067 m$

total range (horizontal)$=r+R=10367.61 m=10.367 km$