Answer:
124.51 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 49.4 m/s
Final velocity (v) = 0 m/s (at maximum height)
Maximum height (h) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:
v² = u² – 2gh ( since the ball is going against gravity)
0² = 49.4² – (2 × 9.8 × h)
0 = 2440.36 – 19.6h
Collect like terms
0 – 2440.36 = –19.6h
–2440.36 = –19.6h
Divide both side by –19.6
h = –2440.36 / –19.6
h = 124.51 m
Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m
