A ball is thrown with initial speed of 30 m/s at angle of 45 o Assume ball is th

Question

A ball is thrown with initial speed of 30 m/s at angle of 45 o Assume ball is thrown from ground level and lands at ground level. Time of flight is 4.3 sec, maximum height is 22.95 m and distance from where ball is thrown to where it lands is 91.84 m What is speed of ball at impact? Please show formula and explain reasoning A ball is thrown with initial speed of 30 m/s at angle of 45 o Assume ball is thrown from ground level and lands at ground level. Time of flight is 4.3 sec, maximum height is 22.95 m and distance from where ball is thrown to where it lands is 91.84 m What is speed of ball at impact? Please show formula and explain reasoning

Explanation / Answer

let the vertical component of initial velocity by Uy and horizontal be Ux then Ux = 30cos45 = 21.21m/s Uy = 30sin45 = 21.21m/s time of flight T = 4.3s this means 4.3/2 s is taken to reach maximum height. t = 2.15s and t = 2.15s to reach to ground from maximum height again. v = u+at 0 = 21.21 + a*2.15 => a = -9.865 =>acceleration due to gravity, g = 9.865m/s2 so, Vy = 0 + 9.865*2.15 =>Vy = 21.21m/s Vx = Ux = 21.21m/s so, speed of ball at impact = sqrt(Vx2 + Vy2) = 30m/s

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