Answer:
1. Time of flight = 7.22 s
2. Range = 255.10 m
3. Maximum height = 63.77 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 50 m/s
Angle of projection (θ) = 45°
Time of flight (T) =?
Range (R) =?
Maximum height (H) =?
1. Determination of the time of flight
Initial velocity (u) = 50 m/s
Angle of projection (θ) = 45°
Acceleration due to gravity (g) = 9.8 m/s²
Time of flight (T) =?
T = 2uSine θ /g
T = 2 × 50 Sine 45 / 9.8
T = 100 × 0.7071 /9.8
T = 7.22 s
2. Determination of the range.
Initial velocity (u) = 50 m/s
Angle of projection (θ) = 45°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =?
R = u²Sine 2θ /g
R = 50² × Sine (2×45) /9.8
R = 2500 × Sine 90 /9.8
R = 2500 × 1 / 9.8
R = 255.10 m
3. Determination of the maximum height.
Initial velocity (u) = 50 m/s
Angle of projection (θ) = 45°
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (H) =
H = u²Sine²θ /2g
H = 50² × (Sine 45)² / 2 × 9.8
H = 2500 × (0.7071)² / 19.6
H = 63.77 m
Answer:
Explanation:
Given;
initial velocity of projection, u = 50 m/s
angle of projection, θ = 45°
Time spent in the air is the time of flight, given as;
[tex]T = \frac{2usin \theta}{g}\\\\T = \frac{2(50)sin (45)}{9.8}\\\\T = 7.22 \ s[/tex]
The range of the projectile is given by;
[tex]R = \frac{u^2sin(2\theta)}{g}\\\\R = \frac{u^2sin(2*45)}{g}\\\\ R = \frac{u^2sin(90)}{g}\\\\ R = \frac{u^2(1)}{g}\\\\R=R_{max} = \frac{u^2}{g} =\frac{50^2}{9.8} = 255.1 \ m[/tex]
The maximum height of the projectile is given by;
[tex]H = \frac{u^2sin^2\theta}{2g}\\\\H = \frac{(50)^2sin^2(45)}{2*9.8}\\\\H = 63.77 \ m[/tex]
