galileo droped a small cannon ball of the top of the leaning tower of pisa. the tower was 40m high what was the speed of the cannon ball whe

You could also approach this using conservation of energy.  If we assume air resistance is negligible then the potential energy (PE) at the top of the tower gets turned into kinetic energy (KE) at the bottom.

PE = m*g*h   where m = mass, g = gravitational acceleration (I usually use 9.81m/s², but I'll use 9.8m/s² here because that's what Chris used), h = tower height (40m)

KE = (1/2)*m*v² where v is speed when the ball hits the ground.

So:

(1/2)*m*v² = m*9.8*40

The mass cancels, multiply through by 2 and take the square root of both sides to get:

v = √(2*9.8*40) m/s

$${\mathtt{v}} = {\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{40}}}} = {\mathtt{v}} = {\mathtt{28}}$$  m/s

galileo droped a small cannon ball of the top of the leaning tower of pisa. the tower was 40m high what was the speed of the cannon ball when it hit the g round

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We need to first find out how long it took to hit the ground. The position of an object  at any time t is given by

y = -4.9t² +v₀t + h

where y is the position of the object at any time, v₀ is the original velocity and h is the original height in meters

So, in our case, "y" = 0 when the object hits the ground, v₀ = 0 (since the object was dropped, and h = 40     So we have

0 = -4.9t² + 40            And solving this we have

$${\mathtt{0}} = {\mathtt{\,-\,}}{\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40}} = \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{{\mathtt{20}}}{{\mathtt{7}}}}\\ {\mathtt{t}} = {\frac{{\mathtt{20}}}{{\mathtt{7}}}}\\ \end{array} \right\}$$  

= (20/7) seconds

And the velocity at time t =

v = -9.8t

v= -9.8(20/7) = -28 m/s   

Note we have to have the negative because the object is moving downward !!!

Do you need/want a calculus answer?

28m/s