a) Find the vertical velocity "v_y=g\\times t = 10\\times 2 = 20" m/s.
Then the ball speed will be "v =\\sqrt {v _ y ^ 2 + v _ x ^ 2} =\\sqrt {20 ^ 2 + 8 ^ 2 }\\approx 21.54" m/s.
The angle of the ball trajectory in 2 seconds is at an angle to the horizon equal to "\\arctg {\\frac {20} {8} }\\approx 68\\degree" .
b) Since the distance travelled (where "1524" meters "\\approx 5000" feet) is "S =\\frac {g\\times t ^ 2} {2}" , therefore t =\sqrt "{\\frac {2\\times S} {g}} =\\sqrt {\\frac {2\\times 1524} {10} }\\approx17.46" s
Then the vertical velocity is "v_y=10\\times 17.46 = 174.6" m/s
"300" miles/hours "\\approx134.08" m/s.
"v_{total}=\\sqrt{174.6^2+134.08^2}\\approx 220.14" m/s
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