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A stone is thrown from the top of a building upward at an (A) how long does it take the stone to reach the ground? Solution We conceptualize the problem by studying Figure To analyze the problem, let us once again separate mo- tion into two components. The initial x and y components To find t, we can use y f " y i & v yi t & a y t 2 (Eq. 4.9a) with y i " 0, y f " # 45.0 m, a y " # g, and v yi " 10.0 m/s (there is a negative sign on the numerical value of y f because we have chosen the top of the building as the origin): Solving the quadratic equation for t gives, for the positive root, t " To finalize this part, think: Does the negative root have any physical meaning? (B) What is the speed of the stone just before it strikes the ground? Solution We can use Equation 4.8a, v yf " v yi & a y t, with t " 4.22 s to obtain the y component of the velocity just be- Because v xf " v xi " 17.3 m/s, the required speed is To finalize this part, is it reasonable that the y component of 35.9 m/s v f " √ v 2 xf & v 2 yf " √ (17.3) 2 & (#31.4) 2 m/s " v yf " 10.0 m/s # (9.80 m/s 2 )(4.22 s) " #31.4 m/s 4.22 s. # 45.0 m " (10.0 m/s)t # 1 2 (9.80 m/s 2 )t 2 1 2 v yi " v i sin
' i " (20.0 m/s)sin
30.0) " 10.0 m/s v xi " v i
cos
' i " (20.0 m/s)cos
30.0) " 17.3 m/s S E C T I O N 4 . 3 • Projectile Motion 89 Example 4.5 That’s Quite an Arm! What If? What if a horizontal wind is blowing in the same direction as the ball is thrown and it causes the ball to have a x ! 0.500 m/s 2 . Which part of this example, (A) or (B), will have a different answer? Answer Recall that the motions in the x and y directions are We can find the new final horizontal velocity component by using Equation 4.8a: and the new final speed: v f " √ v xf 2 & v 2 yf " √ (19.4) 2 & (#31.4) 2
m/s " 36.9
m/s " 19.4
m/s v xf " v xi & a x t " 17.3
m/s & (0.500
m/s 2 )(4.22
s) ! 45.0 m (0, 0) y x v i = 20.0 m/s θ i = 30.0 ° y f = – 45.0 m x f = ? x f Figure 4.14 (Example 4.5) A stone is thrown from the top of a building. Investigate this situation at the Interactive Worked Example link at http://www.pse6.com. Now if we use Equation 4.9a to write an expression for the y Thus, by comparing the two previous equations, we see that y P " x P
tan ' i # 1 2 gt 2 results. That is, when y P " y T , x P " x T . You can obtain the same result, using expressions for the position vectors for To finalize this problem, note that a collision can result only when where d is the initial elevation of the target above the floor. If v i sin ' i is less than this value, the projectile will strike the floor before reaching v i sin ' i - √ gd
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2 Investigate this situation at the Interactive Worked Example link at http://www.pse6.com. Interactive |