projectile is fired with speed v0 at an angle theta fromthe horizontal. Consider
ID: 1754531 • Letter: P
Question
projectile is fired with speed v0 at an angle theta fromthe horizontal. Consider your advice to an artillery officer whohas the following problem. From his current postition, he mustshoot over a hill of height H at a target on the other side, whichhas the same elevation as his gun. He knows from his accurate mapboth the bearing and the distance R to the target and also that thehill is halfway to the target. To shoot as accurately as possible,he wants the projectile to just barely pass above the hill. Findthe angle theta above the horizontal at which the projectile shouldbe fired. What is the initial speed?Explanation / Answer
Horizontal range R = v ^ 2 sin 2 / g = v ^ 2 [ 2 sin cos ] / g = 2 v ^ 2 sin cos / g -----(1) And maximum height H = v ^ 2 sin2 / 2g sin 2 = 2H g / v ^ 2 sin = [ 2Hg ] / v substitue this value in eq ( 1) we get , R = 2 v ^ 2 { [ 2Hg ] / v } cos / g = 2 v [2Hg ] cos / g from this v = Rg / [ 2 [ 2Hg ] cos ] time of flight T = 2v sin / g from eq( 2) , v sin = [ 2gH ] So, T = 2 [ 2gH ] / g = 2 2* [ H / g ] time of flight T = 2v sin / g from eq( 2) , v sin = [ 2gH ] So, T = 2 [ 2gH ] / g = 2 2* [ H / g ]Related Questions
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