Answer:
v = 46.99 m/s
Explanation:
The velocity of the ball just before it touches the ground, is given by the following formula:
(1)
vx: horizontal component of the velocity
vy: vertical component of the velocity
The vertical component vy is calculated by using the following formula:
(2)
vy: final velocity
voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)
g: gravitational acceleration = 9.8 m/s^2
h: height = 1.60m
You replace the values of the parameters in the equation (2):
![v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}](https://tex.z-dn.net/?f=v_y%3D2%289.8m%2Fs%5E2%29%281.60m%29%3D31.36%5Cfrac%7Bm%7D%7Bs%7D)
vx is calculated by using the information about the horizontal range of the ball:
(3)
R: horizontal range of the ball = 20.0 m
You solve the previous equation for vo, the initial horizontal velocity:
![v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}](https://tex.z-dn.net/?f=v_o%3DR%5Csqrt%7B%5Cfrac%7Bg%7D%7B2h%7D%7D%3D%2820.0m%29%5Csqrt%7B%5Cfrac%7B9.8m%2Fs%5E2%7D%7B2%281.60m%29%7D%7D%5C%5C%5C%5Cv_o%3D35%5Cfrac%7Bm%7D%7Bs%7D)
The horizontal component of the velocity is constant in the complete trajectory, hence, you have that
vx = vo = 35 m/s
Finally, you replace the values of vx and vy in the equation (1):
![v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%2835m%2Fs%29%5E2%2B%2831.36m%2Fs%29%5E2%7D%3D46.99%5Cfrac%7Bm%7D%7Bs%7D)
The velocity of the ball just before it touches the ground is 46.99 m/s