Answer:
1. Range = 153.42 m.
2. Maximum height = 54.78 m
3. Time of flight = 6.69 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 ms¯¹
Angle of projection (θ) = 55°
1. Determination of the range.
Initial velocity (u) = 40 ms¯¹
Angle of projection (θ) = 55°
Acceleration due to gravity (g) = 9.8 ms¯²
Range (R) =?
The range i.e how far the ball went can be obtained as follow:
R = u² Sine 2θ /g
R = 40² × Sine (2×55) / 9.8
R = 1600 × Sine 110 / 9.8
R = 1600 × 0.9397 /9.8
R = 153.42 m
2. Determination of the maximum height.
Initial velocity (u) = 40 ms¯¹
Angle of projection (θ) = 55°
Acceleration due to gravity (g) = 9.8 ms¯²
Maximum height (H) =?
H = u² Sine² θ / 2g
H = 40² × (Sine 55)² / 2 × 9.8
H = 1600 × (0.8192) ² / 19.6
H = 54.78 m
3. Determination of the time of flight
Initial velocity (u) = 40 ms¯¹
Angle of projection (θ) = 55°
Acceleration due to gravity (g) = 9.8 ms¯²
Time of flight (T) =?
T = 2u Sine θ / g
T = 2 × 40 × Sine 55 / 9.8
T = 80 × 0.8192 / 9.8
T = 6.69 s
