Answer:
50.2 m/s
Explanation:
First of all, we need to find the time it takes for the ball to reach the ground.
The vertical position of the ball at time t is given by
![y(t) = h +ut + \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20%2But%20%2B%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
h = 120 m is the initial height
u = 0 is the initial vertical velocity
g = -9.8 m/s^2 is the acceleration of gravity
The ball reaches the ground when y = 0. Substituting into the equation and solving for t, we find the time of light:
![0=h+\frac{1}{2}gt^2\\t=\sqrt{-\frac{2h}{g}}=\sqrt{-\frac{2(120)}{-9.8}}=4.95 s](https://tex.z-dn.net/?f=0%3Dh%2B%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5Ct%3D%5Csqrt%7B-%5Cfrac%7B2h%7D%7Bg%7D%7D%3D%5Csqrt%7B-%5Cfrac%7B2%28120%29%7D%7B-9.8%7D%7D%3D4.95%20s)
The vertical component of the velocity of the ball changes following the equation
![v_y(t) = u+gt](https://tex.z-dn.net/?f=v_y%28t%29%20%3D%20u%2Bgt)
Substituting t = 4.95 s, we find the final vertical velocity of the ball just before reaching the ground:
![v_y=0+(-9.8)(4.95)=-48.5 m/s](https://tex.z-dn.net/?f=v_y%3D0%2B%28-9.8%29%284.95%29%3D-48.5%20m%2Fs)
where the negative sign means the direction is downward.
We also can find the horizontal component of the velocity: since we know the horizontal distance travelled is d = 64 m,
![v_x = \frac{d}{t}=\frac{64}{4.95}=12.9 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D%3D%5Cfrac%7B64%7D%7B4.95%7D%3D12.9%20m%2Fs)
And the final speed is calculated as the magnitude of the resultant of the two components of the velocity:
![v=\sqrt{v_x^2+v_y^2}=\sqrt{12.9^2+(-48.5)^2}=50.2 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%3D%5Csqrt%7B12.9%5E2%2B%28-48.5%29%5E2%7D%3D50.2%20m%2Fs)