A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what height h was the ball thrown?

Assume no air resistance, and g = 9.8 m/s².

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

Answer: h = 110.4 m