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Snowcat [4.5K]
3 years ago
8

Could I get help on this question please

Physics
1 answer:
Zepler [3.9K]3 years ago
3 0

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

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