9-10-08LectPHY2048

# 9-10-08LectPHY2048 - Projectile motion examples Centripetal...

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Projectile motion examples 9-10-08 Centripetal acceleration Centripetal acceleration examples

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Projectile motion example 1 A ball is thrown up at some angle with the horizontal. Two seconds after launch it is at a point 40 m horizontally and 53 m vertically. What were the horizontal and vertical components of its launch velocity?
v xo Projectile motion example 1 A ball is thrown up at some angle with the horizontal. Two seconds after launch it is at a point 40 m horizontally and 53 m vertically. What were the horizontal and vertical components of its launch velocity? Sketch the circumstance: v yo v y v x v G r x (2) = 40 m r y (2) = 53 m The position vector of the ball is given by: xy ˆˆ r(t) r (t)i r (t) j =+ G ˆ j ˆ i at time t = 2 s, r x (2) = 40 m r y (2) = 53 m

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v xo v yo v y v x v G r x (2) = 40 m r y (2) = 53 m ˆ j ˆ i Along the vertical axis: 2 yy o y o 1 r( t) r v t gt 2 =+ 0 Solve for v yo , 2 y y yo 1 t) g t 1 2 v gt tt 2 + == + 2 53 m 1 m m (9.8 )(2 s) 36.3 2s 2 s s = Along the horizontal axis there is no acceleration so: xx o x o 0 x xo t) 40m m v2 0 t2 s s = Use r y (2 s) = 53 m ,
A ball is thrown horizontally from the top of a 20 m high cliff. It strikes the ground at an angle of 45 o . With what speed was it thrown? Projectile motion example 2

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A ball is thrown horizontally from the top of a 20 m high cliff. It strikes the ground at an angle of 45 o . With what speed was it thrown? 20 m Since strikes ground at the angle of 45 o : xf yf vv = 22 y ff o 2 g ( y y ) = −− Now, So, 2 mm v (0) 2(9.8 )(0 m 20 m) 392 ss =− = Taking the sq root, xf xo m v1 9 . 8 v v s == = Projectile motion example 2
The basketball is released at 55 o wrt the horizontal. The labeled distances are h 1 = 2.2 m , h 2 = 3.05 m , and d = 4 m . At what speed must the ball be launched to make the basket? For the vertical direction, h 1 h 2 d ˆ i ˆ j 2 1 yy o y o y 2 rr v t a t =+ + Take the floor below the release point as the coordinate origin. Then r yo = h 1 . Also a y = -g so when the ball is at the basket, 2 1 21y o b b 2 hh v t g t For the horizontal direction, , but r xo = 0 and a x = 0 so, 2 1 xx o x o x 2 v t a t = ++ o rv t = Projectile motion example 3

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h 1 h 2 d ˆ i ˆ j h 1 = 2.1 m, h 2 = 3.05 m, d = 4 m, θ o = 55 o xo b dv t = When the ball is at the basket Solving this for the time when the ball is at the basket, b xo d t v = Using this in the y equation at that time, 2 1 21 y o 2 xo xo dd hhv g vv ⎛⎞ −= ⎜⎟ ⎝⎠
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## This note was uploaded on 08/25/2011 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.

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9-10-08LectPHY2048 - Projectile motion examples Centripetal...

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