Question

Activity 1: How far can a soccer player kick a soccer ball down field? Through the application of a linear function and a quadratic function and ignoring wind and air resistance one can describe the path of a soccer ball. These functions depend on two elements that are within the control of the player: velocity of the kick (v k ) and angle of the kick (?). A skilled high school soccer player can kick a soccer ball at speeds up to 50 to 60 mi/h, while a veteran professional soccer player can kick the soccer ball up to 80 mi/h. Vectors Gravity The vectors identified in the triangle describe the initial velocity of the soccer ball as the combination of a vertical and horizontal velocity. The constant g represents the acceleration of any object due to Earth’s gravitational pull. The value of g near Earth’s surface is about ?32 ft/s2 . v x = v k cos ? & v y = v k sin ?

1. Use the information above to calculate the horizontal and vertical velocities of a ball kicked at a 35° angle with an initial velocity of 60 mi/h. Convert the velocities to ft/s. (2 pts) Project 2 368 MTHH 039 2. The equations x(t) = v x t and y(t) = v y t + 0.5 gt2 describe the x- and y- coordinate of a soccer ball function of time. Use the second to calculate the time the ball will take to complete its parabolic path. (4 pts) 3. Use the first equation given in Question 2 to calculate how far the ball will travel horizontally from its original position. (2 pts)

Activity 2: How far can a soccer player kick a soccer ball down field? Through the application of a linear function and a quadratic function and ignoring wind and air resistance one can describe the path of a soccer ball. These functions depend on two elements that are within the control of the player: velocity of the kick (v k ) and angle of the kick (?). A skilled high school soccer player can kick a soccer ball at speeds up to 50 to 60 mi/h, while a veteran professional soccer player can kick the soccer ball up to 80 mi/h.

1. Use the technique developed in Activity 1 to calculate horizontal distance of the kick for angle in 15° increments from 15° to 90°? Make a spreadsheet for your calculations. Use the initial velocity of 60 mi/h. (8 pts)

I want to know the answer of the last question that I write bold and italic. Let me know the answer of this questions!!!

Answer 1

Answer to the last question in bold letters would be as follows:

First of all 60 mi/h = 60*1760*3/60² ft/s = 88ft/sec. Value of g would be taken to be 32ft/s²

v_k is the velocity of kick= 88ft/s. If the angle of kick is 15^o , horizontal component of velocity would be 88 cos15 ft/s and vertical component would be 88 sin15 ft/s.

Due to earth's gravity, the vertical velocity would continue to decrease till the ball reaches its maximum height. The vertcal distance travelled y(t) , in any given time 't' is given by the eq, y(t)= v_y t +1/2 gt² . Since direction of vertical velocity is up ward and the acceleration due to gravity works down wards,taking upward motion as positive, y(t)= v_y t -1/2 gt² . Differentiating this eq w.e.t 't' it would be y'(t) = v_y -gt. y'(t) gives the velocity at any time 't'

The upward velocity would conitue to decrease as stated above, at the maximim height this velocity would become 0. Thus , when y'(t)=0. This gives v_y=gt -> t= v_y /g =(1/32) 88 sin15=(11/4) sin15 seconds. This is the time for the ball to reach the maximum height.

The time taken by the ball to fall back on the ground would also be equal. Hence total time of the flight=(11/2) sin15 seconds.

Now during this whole time the horizontal velocity v_x would remain constant= 88 cos15. Distance travelled x_t in (11/2) sin15 seconds would be = 88 cos15 *(11/2) sin15 ft = 44*11 sin15 cos15 = (22*11) 2sin15 cos15 =22*11 sin30= 22*11*(1/2) =121

Next when the angle of kick is 30^o the horizontal distance travelled would be= (22*11) 2sin30 cos30= (22*11) sin60=121 sqrt3 ft

Next when the angle of kick is 45^o , the distance would be= (22*11) 2sin45 cos45=22*11 sin90 = 242 ft {sin 90^o=1}

Next when the angle of kick is 60^o , the distance would be = (22*11)2sin60cos60= (22*11)sin120= 121 sqrt3 {sin 120=sqrt3 /2)

Next when the angle of kick is 90^o , the distance travelled would be (22*11) 2 sin90 cos90=0     {cos90=0}

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