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A projectile shot at angles of 45^(@) ...

A projectile shot at angles of `45^(@)` above the horizontal strikes a building 30 m away at a point 15 m above the point of projection. Find : (a) the speed of projection . (b) the magnitude and direction of velocity of projectile when it strikes the building .

Text Solution

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Let u is the speed of projection
a. Let P be the point on the building wher projectile hits it.
Taking point of projection as origin coordinates of P are (30,15)
Using the equation of trajectory.
`y=xtan theta-(gx^(2))/(2u^(2)cos^(2)theta)implies15=30tan45^(@)-(g(3)^(2))/(2u^(2)cos^(2)45^(@))impliesu=24.2ms^(-1)`

b. At `p,v_(x)=u_(x)=24.2cos 45^(@)=17.11ms^(-1)`
`v_(y)^(2)=u_(y)^(2)+2a_(y)s_(y)impliesv_(y)^(2)=u^(2)sin^(2)45^(@)-2g(15)`
`v_(y)^(2)=(60g)(0.5)-30g=0impliesv_(y)=0`
At P projectile is at its highest point and hence moving horiontally.
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