8. An object is projected vertically upward with a velocity of 80 ms\(^{-1}\). Find the; a. Maximum height reached (Leave your answer in whole number 'abc.'...

Assessment: WAEC SSCE - Further Mathematics - 2025 (Essay) Subject: Further Mathematics

Question 1 Report

8. An object is projected vertically upward with a velocity of 80 ms\(^{-1}\). Find the;

a. Maximum height reached (Leave your answer in whole number 'abc.')

b. Time taken to return to the point of projection [ g = 10m/s\(^2\)]

Answer

Answer Details

Maximum height, h, v = 0m/s at max height

Recall v\(^2\) = u\(^2\) − 2gh

0\(^2\) = 80\(^2\) − 2(10)h

0 = 6400 - 20h

h = \(\frac{6400}{20}\)

h = 320 m

b. Time taken to reach maximum height, t

V = u − gt

0 = 80−10t

10t = 80

t = \(\frac{80}{10}\) = 8s

Since it takes 8 s to reach maximum height, it will also take 8 s to return to the point of projection; hence, it is 16 s

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