(10 Marks)A long jump is shown in the diagram. The initial velocity vo is 7.0 m/sec and initial angleof release is 45 degrees. Assuming negligible air resistance and acceleration due togravity, g = 10 m/(sec x sec).Also, sin(45)=cos(45)=0.7Select best answer from given choices for each from (a) to (e).(a) What is horizontal component of velocity (Vxh) at the position of maximum heightabove ground?(b) What is the total velocity (Vth) at the position of maximum height above ground?(c) Calculate time (Time) required to reach the maximum height.(d) What is the maximum height (H) achieved?(e) How much total horizontal distance is covered by the athlete?(Hint: time to come back to the ground = 2 x time to reach maximum height)(10 Marks)Select one or more:(a) = 4.90 m/sec(a) = 10.00 m/sec(a) = 8.75 m/sec(b) = 4.90 m/sec(b) = 10.00 m/sec(b) = 6.30 m/sec(c) Time to maximum height = 0.49 sec(c) Time to maximum height = 1.00 sec(c) Time to maximum height = 49.00 sec(d) Maximum height = 1.20 m(d) Maximum height = 5.00 m(d) Maximum height = 480.20 m(e) Horizontal distance = 4.90 m

The horizontal component of velocity at maximum height,ux=u cosθ=7×cos 450=4.90m/sec

Answered: A long jump is shown in the diagram.…

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