Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: −2v0 sin θ −2v sin θ 0 x = (v0 cos θ ) ay and ay Using the fact that 2sin θ cos θ = sin 2θ , we have x=− 2 2v0 cos θ sin θ ay =− 2 v0 sin 2θ ay Thus, we find that sin 2θ = − and x ay 2 v0 =− (91.4 m) (–9.80 m/s 2 ) = 4.91× 10−3 2 (427 m/s) 2θ = 0.281° or 2θ = 180.000° − 0.281° = 179.719° Therefore, θ = 0.141° and 89.860° _____________________________________________________________________________________________ Chapter 3 Problems 131 48. REASONING The angle θ is the angle that the balloon’s initial velocity v0 makes with the horizontal, and can be found from the horizontal and vertical components of the initial velocity: v0 y v0 y (1) tan θ = or θ = tan −1 v v0 x 0x (Note: because the balloon follows a curved trajectory, θ is not related in this fashion to the horizontal and vertical components x and y of the balloon’s displacement.) We will use x = v0 x t + 1 a x t 2 (Equation 3.5a) and y = v0 yt + 1 ayt 2 (Equation 3.5b) to find expressions 2 2 for the horizontal and vertical components (v0x and v0y) of the balloon’s initial velocity, and then combine those results with Equation (1) to find θ. For the initial horizontal velocity component, with ax = 0 m/s2 since air resistance is being ignored, Equation 3.5a gives ( ) x = v0 x t + 1 0 m/s 2 t 2 = v0 x t 2 v0 x = or x t (2) For the initial vertical velocity component, we obtain from Equation 3.5b that y = v0 y t + 1 a t2 2y or v0 y = y − 1 a yt 2 2 t (3) In part b, the initial speed v0 of the second balloon can be found by first noting that it is related to the x component v0x of the balloon’s initial velocity by v0 x = v0 cos θ . Since there is no acceleration in the x direction (ax = 0 m/s2), v0x is equal to the x component of the balloon’s displacement (x = +35.0 m) divided by the time t that the second balloon is in the air, or v0x = x/t. Thus, the relation v0 x = v0 cos θ can be written as x = v0 cosθ t (4) In Equation (4), x = +35.0 m and θ is known from the result of part (a). The time of flight t is related to the y component of the balloon’s displacement by y = v0 yt + 1 ayt 2 (Equation 2 3.5b). Since v0 y = v0 sin θ , Equation 3.5b can be expressed as y = v0 yt + 1 a y t 2 = ( v0 sin θ ) t + 1 a yt 2 2 2 (5) Equations (4) and (5) provide everything necessary to find the initial speed v0 of the water balloon for the second launch. 132 KINEMATICS IN TWO DIMENSIONS SOLUTION a. Substituting Equations (2) for v0x and (3) for v0y into Equation (1) gives an expression for the balloon’s initial direction θ : y − 1 a yt 2 2 y − 1 a yt 2 v0 y t 2 −1 −1 = tan −1 θ = tan = tan x v0 x x t (5) The balloon leaves the roof of Jackson, 15.0 m above the ground, and hits halfway up Walton, a point 1 (22.0 m) = 11.0 m above the ground. The balloon’s height at impact is 2 therefore 15.0 m − 11.0 m = 4.0 m below its launch height. Taking up as the positive direction, the vertical displacement of the balloon is therefore y = −4.0 m, and its horizontal displacement is x = +35.0 m, the distance between the buildings. Therefore, the angle at which the first balloon is launched is ( ) −4.0 m − 1 −9.80 m/s 2 ( 2.0 s )2 2 = 24o θ = tan −1 35.0 m b. The angle θ is that found in part (a), and the vertical displacement is the difference in heights between the Walton & Jackson dorms: y = 22.0 m − 15.0 m = +7.0 m. Solving Equation (4) for the elapsed time t and substituting the result into Equation (5) gives y = ( v0 sin θ ) t + 1 a t2 2y x 1 x = ( v0 sin θ ) + 2 ay v0 cos θ v0 cos θ 2 Solving this equation for the initial speed v0 of the second balloon gives x v0 = cos θ ( ) 1 −9.80 m/s2 35.0 m 2 = = 29 m/s sin θ cos 24° 7.0 m − 35.0 m sin 24° y−x ( ) cosθ cos 24° 1a 2y 49. SSM REASONING Since the horizontal motion is not accelerated, we know that the x component of the velocity remains constant at 340 m/s. Thus, we can use Equation 3.5a (with ax = 0 m/s2) to determine the time that the bullet spends in the building before it is embedded in the wall. Since we know the vertical displacement of the bullet after it enters the building, we can use the flight time in the building and Equation 3.5b to find the y Chapter 3 Problems 133 component of the velocity of the bullet as it enters the window. Then, Equation 3.6b can be used (with v0 y = 0 m/s) to determine the vertical displacement y of the bullet as it passes between the buildings. We can determine the distance H by adding the magnitude of y to the vertical distance of 0.50 m within the building. Once we know the vertical displacement of the bullet as it passes between the buildings, we can determine the time t1 required for the bullet to reach the window using Equation 3.4b. Since the motion in the x direction is not accelerated, the distance D can then be found from D = v0 xt1 . SOLUTION Assuming that the direction to the right is positive, we find that the time that the bullet spends in the building is (according to Equation 3.5a) t= x v0 x = 6.9 m = 0.0203 s 340 m/s The vertical displacement of...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online