{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

96_PartUniversity Physics Solution

# 96_PartUniversity Physics Solution - 3-26 Chapter 3 SET UP...

This preview shows pages 1–3. Sign up to view the full content.

3-26 Chapter 3 S ET U P : x -component : 0 40.0 m, x x = 0, x a = 0 0 cos53.0 x v v = ° 2 1 0 0 2 x x x x v t a t = + E XECUTE : 0 40.0 m ( )cos53.0 v t = ° The second equation says 0 40.0 m 66.47 m. cos53.0 v t = = ° Use this to replace 0 v t in the first equation: 2 2 15.0 m (66.47 m)sin53 (4.90 m/s ) t = ° − 2 2 (66.46 m)sin53 15.0 m 68.08 m 3.727 s. 4.90 m/s 4.90 m/s t ° + = = = Now that we have t we can use the x -component equation to solve for 0 : v 0 40.0 m 40.0 m 17.8 m/s. cos53.0 (3.727 s)cos53.0 v t = = = ° ° E VALUATE : Using these values of 0 v and t in the 1 0 0 2 y y y y v a t 2 = = + equation verifies that 0 15.0 m. y y = − (b) I DENTIFY : 0 (17.8 m/s)/2 8.9 m/s v = = This is less than the speed required to make it to the other side, so he lands in the river. Use the vertical motion to find the time it takes him to reach the water: S ET U P : 0 100 m; y y = − 0 0 sin53.0 7.11 m/s; y v v = + ° = 2 9.80 m/s y a = − 2 1 0 0 2 y y y y v t a t = + gives 100 7.11 4.90 t t 2 = E XECUTE : 2 4.90 7.11 100 0 t t = and ( ) 2 1 9.80 7.11 (7.11) 4(4.90)( 100) t = ± 0.726 s 4.57 s t = ± so 5.30 s. t = The horizontal distance he travels in this time is 0 0 0 ( cos53.0 ) (5.36 m/s)(5.30 s) 28.4 m. x x x v t v t = = ° = = He lands in the river a horizontal distance of 28.4 m from his launch point. E VALUATE : He has half the minimum speed and makes it only about halfway across. 3.64. I DENTIFY : The rock moves in projectile motion. S ET U P : Let y + be upward. 0 x a = , y a g = − . Eqs.(3.22) and (3.23) give x v and y v . E XECUTE : Combining equations 3.25, 3.22 and 3.23 gives 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 cos ( sin ) (sin cos ) 2 sin ( ) v v v gt v v gt gt α α α α α = + = + + . 2 2 2 2 0 0 0 0 1 2 ( sin ) 2 2 v v g v t gt v gy α = = , where Eq.(3.21) has been used to eliminate t in favor of y . For the case of a rock thrown from the roof of a building of height h , the speed at the ground is found by substituting y h = − into the above expression, yielding 2 0 2 v v gh = + , which is independent of 0 α . E VALUATE : This result, as will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y , positive, negative or zero, as long as 2 0 2 0 v gy > . 3.65. I DENTIFY and S ET U P : Take y + to be upward. The rocket moves with projectile motion, with 0 40.0 m/s y v = + and 0 30.0 m/s x v = relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity. E XECUTE : (a) 0 y v = (at maximum height), 0 40.0 m/s, y v = + 2 9.80 m/s , y a = − 0 ? y y = 2 2 0 0 2 ( ) y y y v v a y y = + gives 0 81.6 m y y = (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. (c) Use the vertical motion of the rocket to find the time it is in the air. 0 40 m/s, y v = 2 9.80 m/s , y a = − 40 m/s, y v = − ? t = 0 y y y v v a t = + gives 8.164 s t = Then 0 0 (30.0 m/s)(8.164 s) 245 m. x x x v t = = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Motion in Two or Three Dimensions 3-27 (d) Relative to the ground the rocket has initial velocity components 0 30.0 m/s x v = and 0 40.0 m/s, y v = so it is traveling at 53.1 ° above the horizontal.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}