How to solve these five questions of distance and speed?

1.

The lower arch of the Sydney Harbour Bridge is 503 metres in width and 118 metres high. It is also a parabola. If the zeros of the parabola are at a; = 0 and a; = 503, then the vertex is at the point C] . Hence the equation of the parabola is y = E B- We will use this equation in the next few questions. Note: the Maple syntax for the point [1, 2] is [1 , 2]. Suppose a aming ball is set to re from [0, 0] and xed to the rail of the lower arch of the Sydney Harbour Bridge (ie the trajectory of the ball follows the rail). The ball must reach the point [503,0] in exactly 6 seconds. The a: and y coordinates can be specied as a function oft in seconds. as -m= Sm a: Note: We are assuming a constant m-velocity, ie E should be constant. The distance travelled by the ball at time u seconds is given by the formula u day 2 dy 2 = dt. 3\"\") f (on) +(dt) 0 Hence after5 seconds the ball has travelled (to the nearest metre) metres. Note: you will probably want to use Maple of WolframAlpha to calculate this integral. The speed of the flaming ball is the rate of change in distance travelled. Hence the speed of the flare at time u seconds is: v(u) = 301\") = (m'(u))2 + (y'(u))2. . After 5 seconds the ball is moving at a speed of metres per second (to the nearest integer). . The initial speed of the ball is metres per second (to the nearest integer). . . . . . 472 503 2 Recall: that the equation describing the lower arch of the bridge Is y = 253009 (a: T) + 118. The average height above launch level of the ball is (to the nearest metre) metres. Recall the formula for the speed at time u seconds is given by: v ( 2 ) = any s( 2) = 2' ( 24 ) ) 2 + ( 2 ' ( 26 ) ) 2 . To accommodate even more fireworks in the 2023 schedule, the artistic director requires that the ball must now take only 3 seconds to exactly follow the trajectory of the lower arch. Hence the new initial speed is (to the nearest integer): U(0 = + metres per second