Two-Dimensional Motion and Vectors Problem E

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1 Two-Dimensional Motion and Vectors Problem E PROJECTILES LAUNCHED AT AN ANGLE PROBLEM SOLUTION 1. DEFINE. PLAN The narrowest strait on earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The strait is only about 6.0 m wide. Suppose an athlete wanting to jump over the sea leaps at an angle of 35 with respect to the horizontal. What is the minimum initial speed that would allow the athlete to clear the gap? Neglect air resistance. Given: Unknown: v i =? Diagram: x = 6.0 m = 35 v θ = 35 x = 6.00 m Choose the euation(s) or situation: The horizontal component of the athlete s velocity, v x, is eual to the initial speed multiplied by the cosine of the angle,, which is eual to the magnitude of the horizontal displacement, x, divided by the time interval reuired for the complete jump. v x = v i cos = x t At the midpoint of the jump, the vertical component of the athlete s velocity, v y, which is the upward vertical component of the initial velocity, v i sin, plus the component of velocity due to free-fall acceleration, euals zero. The time reuired for this to occur is half the time necessary for the total jump. v y = v i sin + a y t = 0 v i sin = a y t Rearrange the euation(s) to isolate the unknown(s): Express t in the second euation in terms of the displacement and velocity component in the first euation. v i sin = a y x v i cos ay x v i = si n cos a v i = y x si n cos Problem E 5

2 3. CALCULATE Substitute the values into the euation(s) and solve: Select the positive root for v i. ( 9.81 m/s 6.0 m) (sin 35 cos 35 ) v i = = 7.9 m/s 4. EVALUATE By substituting the value for v i into the original euations, you can determine the time for the jump to be completed, which is 0.9 s. From this, the height of the jump is found to eual 1.0 m. ADDITIONAL PRACTICE 1. In 1993, Wayne Brian threw a spear a record distance of 01.4 m. (This is not an official sport record because a special device was used to elongate Brian s hand.) Suppose Brian threw the spear at a 35.0 angle with respect to the horizontal. What was the initial speed of the spear?. April Moon set a record in flight shooting (a variety of long-distance archery). In 1981 in Utah, she sent an arrow a horizontal distance of m. What was the speed of the arrow at the top of the flight if the arrow was launched at an angle of 45.0 with respect to the horizontal? 3. In 1989 during overtime in a high school basketball game in Erie, Pennsylvania, Chris Eddy threw a basketball a distance of 7.5 m to score and win the game. If the shot was made at a 50.0 angle above the horizontal, what was the initial speed of the ball? 4. In 1978, Geoff Capes of the United Kingdom won a competition for throwing 5 lb bricks; he threw one brick a distance of 44.0 m. Suppose the brick left Capes hand at an angle of 45.0 with respect to the horizontal. a. What was the initial speed of the brick? b. What was the maximum height reached by the brick? c. If Capes threw the brick straight up with the speed found in (a), what would be the maximum height the brick could achieve? 5. In 1991, Doug Danger rode a motorcycle to jump a horizontal distance of 76.5 m. Find the maximum height of the jump if his angle with respect to the ground at the beginning of the jump was Michael Hout of Ohio can run meter hurdles in 18.9 s at an average speed of 5.8 m/s. What makes this interesting is that he juggles three balls as he runs the distance. Suppose Hout throws a ball up and forward at twice his running speed and just catches it at the same level. At what angle,,must the ball be thrown? (Hint: Consider horizontal displacements for Hout and the ball.) 6 Holt Physics Problem Workbook

3 7. A scared kangaroo once cleared a fence by jumping with a speed of 8.4 m/s at an angle of 55. with respect to the ground. If the jump lasted 1.40 s, how high was the fence? What was the kangaroo s horizontal displacement? 8. Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea s initial speed is. m/s, and that it leaps at an angle of 1 with respect to the horizontal. If the jump lasts 0.16 s, what is the magnitude of the flea s horizontal displacement? How high does the flea jump? Problem E 7

4 Givens 8. y = 1.95 m v x = 3.0 m/s Solutions v y = a y y v = v x + v y = a v x + y y v = (3.0 /s) m ( m/s m) v = 9. 0m s / m / = tan 1 vy v x s = 47.3 m s / = 6.88 m/s ( 9.81 s m/.95 1m) 3.0 m/s = tan 1 a y y = tan 1 vx = 64 below the horizontal Additional Practice E 1. x = 01.4 m = 35.0 y = v i (sin ) t + 1 a y( t) = v i (sin ) + 1 a y t = 0 x = v i (cos ) t x t = vi (cos ) v i (sin ) = 1 y a x v i (cos) v i = a (sin y x cos = ) ( 9.81 m/s 01.4 m) (sin 35.0 cos 35.0 ) v i = 45.8 m/s II Copyright by Holt, Rinehart and Winston. All rights reserved.. x = m = x = 7.5 m = x = 44.0 m = 45.0 v i = a (sin y x cos ) m) (sin 45.0 cos 45.0 ) v i = 96.5 m/s At the top of the arrow s flight: v = v x = v i (cos ) = (96.5 m/scos 45.0 ) = 68. m/s v i = a (sin y x cos ) 7.5 m) (sin 50.0 cos 50.0 ) v i = 16.6 m/s a. v i = a (sin y x cos ) 44.0 m) (sin 45.0 cos 45.0 ) v i = 0.8 m/s Section Two Problem Workbook Solutions II Ch. 3 7

5 Givens Solutions b. At maximum height, v y, f = 0 m/s v y, f = v y, i + a y y max = 0 v y max = = v i ( sin ) (0.8 m/s) (sin 45.0 ) y,i = = 11.0 m ay ( 9.81 m/s ay ) The brick s maximum height is 11.0 m. c. y max = vy, i ( 0. 8 m/s) = ay ( m/ s =.1 m ) The brick s maximum height is.1 m. II 5. x = 76.5 m = 1.0 At maximum height, v y, f = 0 m/s. v y, f = v y, i + a y y max = 0 y max = vy, i = v i ( sin ) ay ay Using the derivation for v i from problem 1, a y max = y x (sin cos ) (s in ) = x(si a y 4(co n ) s ) = x(t an ) 4 y max = (76.5 m tan 1.0 ) = 4.07 m 4 6. v runner = 5.8 m/s v i,ball = v runner In x-direction, v i,ball (cos ) = v runner (cos ) = v runner (cos ) = 1 = cos 1 1 = v i = 8.4 m/s = 55. t = 1.40 s 8. v i =. m/s = 1 t = 0.16 s For first half of jump, t 1 = 1.4 0s = s y = v i (sin ) t a y ( t 1 ) = (8.4 m/ssin s) + 1 ( 9.81 m/s s) y = 4.84 m.40 m =.44 m The fence is.44 m high. x = v i (cos ) t x = (8.4 m/s) (cos s) = 6.73 m x = v i (cos ) t = (. m/s) (cos s) = 0.33 m Maximum height is reached in a time interval of t y max = v i (sin ) t + 1 a y t y max = (. m/ssin 1 ) 0.1 6s + 1 ( 9.81 m/s ) 0.1 6s Copyright by Holt, Rinehart and Winston. All rights reserved. y max = m m = m = 3. cm The flea s maximum height is 3. cm. II Ch. 3 8 Holt Physics Solution Manual

Bell Ringer. x- direction: Ball and car start with same position and velocity, a=0, so always have same position

Bell Ringer. x- direction: Ball and car start with same position and velocity, a=0, so always have same position Objectives Students should be able to add, subtract, and resolve displacement and velocity vectors so they can: Determine the components of a vector along two specified, mutually perpendicular axes. Determine