A projectile fired vertically up from a cannon rises 200 m before returning to the ground. If the same cannon were to fire the same projectile at an angle 45 degrees, then the maximum rangerange would be m.
Asked twice; answered once
To find the maximum range of a projectile fired at an angle of 45 degrees, we can use the formula for the range of a projectile:
Range = (Initial velocity^2 * sin(2θ))/g
Where:
- Initial velocity is the speed at which the projectile is launched
- θ is the angle at which the projectile is launched
- g is the acceleration due to gravity (-9.8 m/s^2)
We are given that the projectile rises 200 m before returning to the ground when fired vertically. Since the vertical displacement is the same as the maximum height reached, we can use this information to find the initial velocity.
Let's break down the vertical motion first. The time taken for the projectile to reach the maximum height can be found using the formula:
Vertical displacement = (Initial velocity * time) + (0.5 * g * time^2)
Rearranging the equation, we get:
t = (Vf - Vi) / g
Since the projectile is at rest at its maximum height, the final velocity (Vf) is 0. Substituting this value into the equation gives us:
t = -Vi / g
We are given that the vertical displacement is 200 m. Substituting this value into the equation, we get:
200 = (Vi * t) + (0.5 * g * t^2)
Simplifying the equation further, we get:
100 = Vi * t
Now we can substitute the value of t into the equation:
100 = Vi * (-Vi / g)
Simplifying further, we get:
100 = -Vi^2 / g
Vi^2 = -100g
Vi = sqrt(-100g)
Vi ≈ 31.62 m/s
Now that we have the initial velocity, we can use the formula for range to find the maximum range when the projectile is fired at an angle of 45 degrees:
Range = (Vi^2 * sin(2θ))/g
Range = (31.62^2 * sin(90))/9.8
Range ≈ 101.4 m
Therefore, if the same cannon were to fire the same projectile at an angle of 45 degrees, the maximum range would be approximately 101.4 meters.