Answer:
A. 10.84 m/s
B. 1.56 s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 45°
Maximum height (H) = 3 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity of projection (u) =?
Time (T) taken to hit the ground again =?
A. Determination of the velocity of projection.
Angle of projection (θ) = 45°
Maximum height (H) = 3 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity of projection (u) =?
H = u²Sine²θ / 2g
3 = u²(Sine 45)² / 2 × 9.8
3 = u²(0.7071)² / 19.6
Cross multiply
3 × 19.6 = u²(0.7071)²
58.8 = u²(0.7071)²
Divide both side by (0.7071)²
u² = 58.8 / (0.7071)²
u² = 117.60
Take the square root of both side
u = √117.60
u = 10.84 m/s
Therefore, the velocity of projection is 10.84 m/s.
B. Determination of the time taken to hit the ground again.
Angle of projection (θ) = 45°
Velocity of projection (u) = 10.84 m/s
Time (T) taken to hit the ground again =?
T = 2uSine θ /g
T = 2 × 10.84 × Sine 45 / 9.8
T = 21.68 × 0.7071 / 9.8
T = 1.56 s
Therefore, the time taken to hit the ground again is 1.56 s.
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