A ball is thrown with an initial velocity of 20mls at an angle of 30 to the horizontal calculate time of flight maximum height the horizontal range of the ball (2)A bullet is fired with a velocity of 50meter per seconds at an angle 0 to the horizontal (a)what is the value of 0 for maximum range with this velocity (b) calculate the maximum range
1. See Previous post: Mon, 10-24-16, 5:42 PM.
2. Vo = 50m/s
a. A = 45o for max. range.
b. R = Vo^2*sin(2A)/g.
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To solve these problems, we will use fundamental principles of projectile motion. Let's go step by step:
Problem 1:
Given:
- Initial velocity (u) = 20 m/s
- Angle of projection (θ) = 30 degrees
a) Time of Flight:
To calculate the time of flight (t), we need to use the formula: t = 2 * (u * sin(θ))/g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we have:
t = 2 * (20 * sin(30))/9.8
Calculating further, we get:
t ≈ 4.082 seconds
Therefore, the time of flight is approximately 4.082 seconds.
b) Maximum Height:
To find the maximum height (H), we use the formula: H = (u² * (sin(θ))²) / (2g)
Plugging in the given values, we have:
H = (20² * (sin(30))²) / (2 * 9.8)
Simplifying further, we get:
H ≈ 6.12 meters
Therefore, the maximum height is approximately 6.12 meters.
c) Horizontal Range:
To calculate the horizontal range (R), we use the formula: R = u * t * cos(θ)
Given:
u = 20 m/s
t = 4.082 seconds
θ = 30 degrees
Plugging these values into the formula, we have:
R = 20 * 4.082 * cos(30)
Calculating further, we get:
R ≈ 70.7 meters
Therefore, the horizontal range is approximately 70.7 meters.
Problem 2:
Given:
- Initial velocity (u) = 50 m/s
- Angle of projection (θ) = 0 degrees (horizontal)
a) Value of θ for Maximum Range:
For maximum range, the angle of projection (θ) is 45 degrees. Therefore, θ = 45 degrees in this case.
b) Maximum Range:
To calculate the maximum range (R), we will use the formula: R = (u² * sin(2θ)) / g
Given:
u = 50 m/s
θ = 45 degrees
Plugging in the values, we have:
R = (50² * sin(2 * 45)) / 9.8
Calculating further, we get:
R ≈ 510.2 meters
Therefore, the maximum range is approximately 510.2 meters.