theExpertTA.com Class Management | Help | Student: telghazi2101@student.stcc Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (6%) Problem 4: A car moves along a horizontal road with constant velocity - (33.6 m/s) i L until it encounters a smooth inclined hill. It climbs the hill with constant velocity U1 - V1,z i + V1,y j - (34.2 m/s) i + (6.23 m/s) 3 as indicated in the figure. The period of time during which the car changes its velocity is At = 3.19 s. *50% Part (a) Enter an expression, in Cartesian unit-vector notation, for the average acceleration of the car during the given time period using the symbols provided. dave = Grade Summary Deductions Potential 100 At 1 () 7 8 9 HOME Submissions d g 4 5 6 Attempts remaining: 9 (0% per attempt) m Vo 1 2 3 detailed view V1 END 10% VOX Vix 0 0% Vly VO LEAR 0% 0% Submit Hint Feedback I give up! Hints: 1 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback -What is the definition of average acceleration? Make sure you enter this as a vector expression, which means you need to understand what the unit vectors are. 50% Part (b) What is the magnitude, in meters per squared second, of the car's average acceleration during the given time period. davg = 1.962 m/s2 Correct! All content @ 2023 Expert TA, LLCClass Management | Help theExpertTA.com | Student: telghazi2101@student.stcc.edu Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (10%) Problem 9: A student standing on a cliff throws a projectile from a vertical height of d = 8.0 m above the level ground with velocity vo = 19 m/s at an angle 0 = 320 below the horizontal, as shown. It moves without air resistance. Use a Cartesian coordinate system with the origin at the initial position of the projectile. To 50% Part (a) With what speed, in meters per second, does the stone strike the ground? m/s Grade Summary Deductions . 0% Potential 100% sin() cos() tan() TT ( 7 8 9 HOME Submissions cotan() asin() acoso 4 5 6 Attempts remaining: 997 (0% per attempt) atan() acotan() sinh() 1 2 3 detailed view tanh() 0 END 0% cosho cotanh() 10% D O Degrees O Radians VO CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 6 Feedback: 0% deduction per feedback 50% Part (b) If the projectile had been thrown from the clifftop with the same initial speed and the same angle, but above the horizontal, would its impact velocity be different?Class Management | Help theExpertTA.com Student: telghazi2101@student.sto Week3 Begin Date: 6/15/2023 11:59:00 PM -- Due Date: 6/27/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM (10%) Problem 10: During service, a tennis ball is struck at height h - 2.24 m above a tennis court, and it moves with the initial velocity vo = (27.7 m/s) i. Assume the ball encounters no air resistance. Use a Cartesian coordinate system with the origin located at the ball's initial position, with i directed horizontally, parallel to the court's surface, and j directed vertically upwards. 33% Part (a) Calculate the horizontal component, in meters per second, of the ball's velocity as it strikes the court. Uf,I m/s Grade Summary Deductions 0% Potential 100% sin( cos( ) tan() It ( ) 7 8 9 HOME Submissions cotan() asin() acos() E 4 5 6 Attempts remaining 998 (0% per attempt) atan( ) acotan() sinh() 1 2 3 detailed view cosh() tanh() cotanh() 0 0% O Degrees O Radians VO BACKSPA CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 feedback: 0% deduction per feedback 33% Part (b) How long, in seconds, does the tennis ball remain in the air before striking the court? tf = 0.6750 s ~ Correct! $ 33% Part (c) What horizontal distance, in meters, does the tennis ball travel before striking the court? If = 15.20 m X Attempts Remain All content @ 2023 Expert TA, LLCProblem;- 04 , Parti@ Here Given , VoLy = 0 = ( 3 3 . 6 m / s ) ? = Initial velocity = ( 34 . 2 m / S ) 2 + ( 6 . 23 m/ s ) ] = final itrest.Velocity At = 3.19s. Now, In order to fund average acceleration, Let a be average acceleration, which can be resolved sim homzontal and Vertical company ent . a = ani + and Now 34. 2 - 33 . 6 2 0 . 188 m / 5 2 3. 19 and Qy (v,ly - Voy ) (6. 23 -0) - 1.953 m/s 2 At 3. 19 G average acceleration will be,- a = (o. 18 8 m / s ?) i + (1 . 9 53 m / s 2 )J Part (b) Magnitude of average acceleration, 1 0 1 = (0 . 188 ) 2 + ( 1 . 953 ) 2 3 . 849 1. 962 m/s2