To solve this problem you use the derived formula ∆dh=(V1h2(sin2Ѳ))/g 250m=V1h2(sin2(17))/9.8m/s2 V1h=((sin(17))/9.8m/s2)1/2 V1h= 64.7292201m/s V1h= 65m/s
The weapon should be fired at a 45-degree angle from the horizontal to achieve the minimum distance traveled by the projectile. This angle maximizes the range (horizontal distance) of the projectile by balancing the vertical and horizontal components of its velocity. At any other angle, the total distance traveled would be greater.
A 45-degree throw maximizes the horizontal distance traveled by balancing the vertical and horizontal components of the projectile's velocity. At this angle, the horizontal component is at its maximum, maximizing the range the object can travel before hitting the ground. Any angle higher or lower will result in a shorter distance traveled.
To solve this problem you use the derived formula ∆dh=(V1h2(sin2Ѳ))/g 250m=V1h2(sin2(17))/9.8m/s2 V1h=((sin(17))/9.8m/s2)1/2 V1h= 64.7292201m/s V1h= 65m/s
its 45 degree
A degree is an angular measure and cannot be measured in millimetres. A 1 degree rise can be interpreted as a ratio of a rise (in millimetres) per a distance of horizontal displacement.
When an object is launched at a 45-degree angle, it splits the initial velocity into horizontal and vertical components equally. This allows for the maximum range because the horizontal component remains constant throughout the flight, while the vertical component decreases due to gravity. This balance between horizontal and vertical components at 45 degrees results in the maximum distance traveled before hitting the ground.
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
Both the arrows will cover equal horizontal distance as their angles are at an difference from 45 degree. Horizontal range is maximum at 45 degrees and decreases equally on sum or difference of an angle from 45. But vertical distance increases on addition of an angle from 45 and decreases on subtraction of angle from it. For more details, contact at saqibahmad81@yahoo.com
The answer is 0.33
Horizontal position is strate 90 degree
It will increase until you hit 45 degrees and the it will begin to drop.
To calculate the vertical drop over a given horizontal distance due to a slope, we use the formula: vertical drop = horizontal distance * tan(slope angle). Given a 3-degree slope over 1 meter, the vertical drop would be 1 meter * tan(3 degrees), which is approximately 0.0524 meters or 5.24 centimeters. This means that for every 1 meter of horizontal distance, the elevation would decrease by about 5.24 centimeters.