Physics Solution Manual for 1100 and 2101

4 ms the speed of the ball just before it strikes the

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Unformatted text preview: particular time t, the speed can be obtained from v = v x + v y . SOLUTION a. Since the ball rolls off the cliff horizontally, v0y = 0. If the origin is chosen at top of the cliff and upward is assumed to be the positive direction, then the vertical component of the ball's displacement is y = – 15.5 m. Thus, Equation 3.5b gives t= 2y = ay 2( −15.5 m) = 1.78 s (–9.80 m / s 2 ) b. Since there is no acceleration in the x direction, v x = v0 x = 11.4 m/s . The y component of the velocity of the ball just before it strikes the water is, according to Equation 3.3b, v y = v0 y + a y t = 0 + (–9.80 m/s2 )(1.78 s) = –17.4 m/s The speed of the ball just before it strikes the water is, therefore, 2 2 v = v x + v y = (11.4 m / s) 2 + ( −17.4 m / s) 2 = 20.8 m / s ____________________________________________________________________________________________ 110 KINEMATICS IN TWO DIMENSIONS 22. REASONING The vehicle’s initial velocity is in the +y direction, and it accelerates only in the +x direction. Therefore, the y component of its velocity remains constant at vy = +21.0 m/s. Initially, the x component of the vehicle’s velocity is zero (v0x = 0 m/s), but it increases at a rate of vx +y +x vy v θ ax = +0.320 m/s2, reaching a final value vx given by the relation v x = v0 x + a x t (Equation 3.3a) when the pilot shuts off the RCS thruster. Once we have found vx, we will use the right triangle shown in the drawing to find the magnitude v and direction θ of the vehicle’s final velocity. SOLUTION a. During the 45.0-second thruster burn, the x component of the vehicle’s velocity increases from zero to v x = v0 x + a x t = ( 0 m/s ) + 0.320 m/s 2 ( 45.0 s ) = 14.4 m/s ( ) Applying the Pythagorean theorem to the right triangle in the drawing, we find the magnitude of the vehicle’s velocity to be: 2 v = vx + v 2 = y (14.4 m/s )2 + ( 21.0 m/s )2 = 25.5 m/s b. Referring again to the drawing, we see that the x and y components of the vehicle’s velocity are related by the tangent of the angle θ. Therefore, the angle of the vehicle’s final velocity is v 14.4 m/s o θ = tan −1 x = tan −1 = 34.4 vy 21.0 m/s 23. SSM REASONING Since the magnitude of the velocity of the fuel tank is given by 2 2 v = v x + v y , it is necessary to know the velocity components vx and v y just before impact. At the instant of release, the empty fuel tank has the same velocity as that of the plane. Therefore, the magnitudes of the initial velocity components of the fuel tank are given by v0 x = v0 cos θ and v0 y = v0 sin θ , where v0 is the speed of the plane at the instant of release. Since the x motion has zero acceleration, the x component of the velocity of the plane remains equal to v0 x for all later times while the tank is airborne. The y component of the velocity of the tank after it has undergone a vertical displacement y is given by Equation 3.6b. Chapter 3 Problems 111 SOLUTION a. Taking up as the positive direction, the velocity components of the fuel tank just before it hits the ground are v x = v0 x = v cos θ = (135 m/s) cos 15° = 1.30 × 102 m/s From Equation 3.6b, we have 2 v y = – v 0 y + 2 a y y = – ( v 0 sin θ ) 2 + 2 a y y =– ( 135 m / s) sin 15.0 ° 2 + 2 ( −9 .80 m / s 2 )( −2.00 × 10 3 m) = –201 m / s Therefore, the magnitude of the velocity of the fuel tank just before impact is 2 2 v = v x + v y = (1.30 × 10 2 m / s) 2 + ( 201 m / s) 2 = 239 m / s The velocity vector just before impact is inclined at an angle φ above the horizontal. This angle is 201 m / s φ = tan −1 = 57.1° 1.30 × 10 2 m / s F G H I J K b. As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the same altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the velocity of the fuel tank in part b just before impact is 239 m/s at an angle of 57.1° above the horizontal . ____________________________________________________________________________________________ 24. REASONING In the absence of air resistance, the horizontal velocity component never changes from its initial value of v0x. Therefore the horizontal distance D traveled by the criminal (which must equal or exceed the distance between the two buildings) is the initial velocity times the travel time t, or D = v0 x t . The time t can be found by noting that the motion of the criminal between the buildings is that of a projectile whose acceleration in the y direction is that due to gravity (ay = −9.80 m/s2, assuming downward to be the negative direction). The relation y = v0 y t + 1 a y t 2 (Equation 3.5b) allows us to determine the time, since y, v0y, and ay are 2 known. Since the criminal is initially running in the horizontal direction, v0y = 0 m/s. Setting v0y = 0 m/s and solving the equation above for t yields t = 2 y / a y . In this result, y = −2.0 m, since downward is the negative direction. 112 KINEMATICS IN TWO DIMENSIONS SOLUTION The horizontal distance traveled after launch is D = v0 x t . Substituting t = 2 y / a y into this rela...
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