a) Find the vertical velocity v y = g × t = 10 × 2 = 20 v_y=g\times t = 10\times 2 = 20 v y = g × t = 10 × 2 = 20 m/s.
Then the ball speed will be v = v y 2 + v x 2 = 2 0 2 + 8 2 ≈ 21.54 v =\sqrt {v _ y ^ 2 + v _ x ^ 2} =\sqrt {20 ^ 2 + 8 ^ 2 }\approx 21.54 v = v y 2 + v x 2 = 2 0 2 + 8 2 ≈ 21.54 m/s.
The angle of the ball trajectory in 2 seconds is at an angle to the horizon equal to arctg 20 8 ≈ 68 ° \arctg {\frac {20} {8} }\approx 68\degree arctg 8 20 ≈ 68° .
b) Since the distance travelled (where 1524 1524 1524 meters ≈ 5000 \approx 5000 ≈ 5000 feet) is S = g × t 2 2 S =\frac {g\times t ^ 2} {2} S = 2 g × t 2 , therefore t = 2 × S g = 2 × 1524 10 ≈ 17.46 t =\sqrt {\frac {2\times S} {g}} =\sqrt {\frac {2\times 1524} {10} }\approx17.46 t = g 2 × S = 10 2 × 1524 ≈ 17.46 s
Then the vertical velocity is v y = 10 × 17.46 = 174.6 v_y=10\times 17.46 = 174.6 v y = 10 × 17.46 = 174.6 m/s
300 300 300 miles/hours ≈ 134.08 \approx134.08 ≈ 134.08 m/s.
v t o t a l = 174. 6 2 + 134.0 8 2 ≈ 220.14 v_{total}=\sqrt{174.6^2+134.08^2}\approx 220.14 v t o t a l = 174. 6 2 + 134.0 8 2 ≈ 220.14 m/s
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