Let x = angle that the initial velocity makes with the horizontal. u = 30 cos(x), horizontal velocity v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain [30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0 55 tan(x) - 16.469 sec²x = 0 55 tan(x) - 16.469[1 + tan²x] = 0 16.469 tan²x - 55 tan(x) + 16.469 = 0 tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula. tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326 Therefore x = 71.6° or x = 18.4°
The time of flight is t = 1.8333 sec(x) = 5.8096 s or 1.932 s The initial vertical velocity is v = 30 sin(x) = 28.467 m/s or 9.468 m/s The horizontal velocity is u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s, u*t = 9.467*5.8096 = 55 m (Correct) or u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is t = 5.8096 s x = 71.6° u = 9.467 m/s v = 28.467 m/s
The height from which the ball was thrown is h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m The ball was thrown from a height of 110.4 m
