
By studying this 2025 WAEC GCE Physics likely questions and answers, you’ll understand the most frequently repeated topics and how to approach each question type in both Paper 1 (Objective) and Paper 2 (Theory).
MA COACH
today
1k+
WAEC GCE, NECO GCE


(ii) I. Acceleration during the first 5s = gradient of line OA
Gradient of OA = 30 / 5 = 6 units
Therefore, during the first 5s, a = 6ms⁻²
II. Deceleration during the last 10s is the gradient of line BC
Gradient of BC = 30 / (30 – 20) = 30 / 10 = 3 units
Therefore, during the last 10s, Deceleration = 3ms⁻²
III. Total distance covered throughout the motion = Area of Trapezium OABC
= (1/2) * [30 + (20 – 5)] * 30
= (1/2) * (30 + 15) * 30
= (1/2) * 45 * 30
= 15 * 45 = 675 units
Therefore, distance = 675m
5) An object is projected into the air with a speed of 50ms⁻¹ at an angle 30° above the ground level. Calculate the maximum height attained by the object [g = 10ms⁻²]
Solution
Given:
H = maximum height = ?
U = speed of projection = 50 ms⁻¹
θ = angle of projection = 30°
g = acceleration of free fall = 10 ms⁻²
The formula for maximum height (H) in projectile motion is:
H = (U² sin²θ) / (2g)
Substitute the given values:
H = (50² × sin²(30°)) / (2 × 10)
H = (2500 × (0.5)²) / 20
H = (2500 × 0.25) / 20
H = 625 / 20
H = 31.25 m
+2349061221656, +2348062853040
admin@macoach.com.ng
+2349061221656, +2348062853040
admin@macoach.com.ng
2nd Floor former joybell nursery & primary school opposite lotogbe junction ondo city, ondo state
Monday — Friday 8am – 11pm
Saturday — 8am – 10pm
Sunday — Closed
MACOACH.com.ng –Your No.1 Online Academy for Daily Mathematics Lessons, Textbook Solutions, and Exam Preparation in Nigeria.

© 2025 Created with MA COACH
