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Unformatted text preview: Physics 1A HW2 Solutions Note: There are many methods to do many of these problems. I give only one for each. 1. Horizontal Cannon on a Cliff (a) The rocket’s vertical position is given by y = H gt 2 / 2, so the time t g it takes for the rocket to hit the ground is found via 0 = H gt 2 g / 2. Thus t g = p 2 H/g and y ( t g / 2) = H g ( t g / 2) 2 / 2 = 3 H/ 4. (b) D is the horizontal distance until landing, so D = v t t ⇒ v = D/t g = D p g/ 2 H . (c) Since horizontal velocity is constant, the rocket is at x = D/ 2 at time t D/ 2 = t g / 2. The rocket’s height at t D/ 2 is that found in part (a): y ( t D/ 2 ) = 3 H/ 4. 2. Kicking a Field Goal (a) Gravity is always conspiring to pull the football downward. If the ball is kicked extremely hard, it would travel in essentially a straight line, and if the kick is aimed anywhere below the bar, it would never make it over. So the minimum angle is φ = tan 1 ( h/d ), where h is the bar height and d is the distance along the ground to the goal post. (b) We know that the angle of the kick is φ =45 ◦ . With the origin set at the kicker, the horizontal position of the ball is x ( t ) = v cos( φ ) t and vertical is y ( t ) = v sin( φ ) t gt 2 / 2. So if t d is defined such that x ( t d ) = d = v cos( φ ) t d , then for the ball to just make it over the bar, we require y ( t d ) = h = v sin( φ ) t d gt 2 d / 2. We now solve for v by first eliminating t d : t d = d/ ( v cos φ ) ⇒ h = tan φ g ( d/ ( v cos φ )) 2 / 2. Solving for v , we have v = d cos φ p 2( d tan φ h ) /g . (c) Convert the answer to part(b), which is in m/s, to km/hr by dividing by 1000 m/km and multiplying by 3600 s/hr....
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 Spring '11
 LIU
 Physics, Acceleration, Circular Motion, Velocity, Cos, /2

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