A ball is thrown upward.What is its maximum height? Its initial vertical speed is 11.6 m/s and the acceleration of gravity is 9.8 m/s 2 . Neglect air resistance. Answer in units of m.

Answer 1

Another way to solve this is to use energy conservation. Set the initial kinetic energy (at moment of launch from the ground at height zero) equal to the potential energy at maximum height h (when the speed is zero), and solve for h.

mv²/2 = mgh

v²/2 = gh

h = v²/(2g) = (11.6²) / [2(9.8)] m = 6.87 m (with roundoff)

Answer 2

If you throw a ball upward, what happens to it? Unless you can throw the ball at 11.2 km/s, it will reach some maximum height then start falling. So how fast is the ball going at the point where it starts falling again? The answer to this question is the key to solving the question.

The ball is at rest at its maximum height ( v = 0).

General equations of motion (neglecting air resistance):

y = y0 + vo * t + (1/2) a * t^2

v = v0 + a*t

Define up as the positive direction and the starting point as y0 = 0. v0 = 11.6 m/s (given) and a = - 9.8 m/s^2 (acceleration due to gravity), v = 0 at the maximum height.

The velocity equation can be solved for time: t = (v - v0) / a

IN this case, t = (- 11.6 m/s)/ (- 9.8 m/s^2) = 1.18 s

Use the time calculated from the velocity equation to find the maximum height:

y = 0 + (11.6 m/s)*(1.18 s) - (1/2) (9.8 m/s^2) (1.18 s)^2

y = 13.7 m - 6.8 m = 6,9 m