A stone is thrown directly upwards with an initial speed of 34.5 m/s . What is the maximum height the stone will reach?

Maximum height is reached when all of the inital kinetic energy is converted to gravitational potential energy. This happens at height H given by

(1/2) M Vo^2 = M g H

H = Vo^2/(2g)

Vo is the initial velocity and g is the accleration of gravity.

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To find the maximum height the stone will reach, you can use the equations of motion. The key equation you'll need is the one that relates the final velocity, initial velocity, acceleration, and displacement:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

In this scenario, the stone is thrown directly upwards, so the final velocity at the maximum height will be zero since the stone momentarily stops before falling back down. The initial velocity is 34.5 m/s, and the acceleration will be the acceleration due to gravity, which is approximately 9.8 m/s^2 (assuming no air resistance). We want to find the displacement, which represents the maximum height.

Rearranging the equation, we have:

0 = (34.5)^2 + 2 * (-9.8) * s

Simplifying this equation, we get:

0 = 1190.25 - 19.6s

Now, solve for s:

19.6s = 1190.25

s = 1190.25 / 19.6

s ≈ 60.77 meters

Therefore, the maximum height the stone will reach is approximately 60.77 meters.