Question #226693

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?

Expert's answer

"\u03b8=25\u00b0 \\\\\nR= u \\cos \u03b8\u00d7T\\\\\nT=total \\space time \\space of \\space flight\\\\\nT=\\frac{2u sin \u03b8}{R}\\\\\nR=u \\cos \u03b8\u00d7 \\frac{2u sin \u03b8}{g} \\\\\nR =\\frac{u^2 sin 2\u03b8}{g}\\\\\nR =\\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\u00d7t+\\frac{1}{2}\u00d79.8\u00d7t^2\\\\\nt=2.497 s"

During this time interval, the extra range covered will be

"r=u cos \u03b8\u00d7t \\\\\n=350 cos 25\u00d72.497 \\\\\n=792.067 m"

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

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