Answer:
16.74 m/s
Explanation:
From the question, the parameters given are:
Height h = 12.2 metres
Initial velocity U = 13.9 m/s
Time t = 1.58 s
The range or distance covered by the ball will be
R = u^2/g
Where g = 9.8 m/s^2
R = 13.9^2/9.8
R = 193.21/9.8
R = 19.72 m
Let first calculate the acceleration by using the formula below
R = Ut + 1/2at^2
Substitute all the parameters into the formula
19.72 = 13.9 × 1.58 - 1/2 × a × 1.58^2
19.72 = 21.962 - 1.2482a
Collect the like terms
-1.2482a = 19.72 - 21.962
- 1.2482a = - 2.242
a = 2.242/1.2482
a = 1.796 m/s^2
Using equation 1 to find the final speed
V = U + at
V = 13.9 + 1.796 × 1.58
V = 13.9 + 2.838
V = 16.74 m/s
Therefore, it hit the ground at speed 16.74 m/s
