Wiki User
∙ 15y agoTo 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
Wiki User
∙ 15y agoThe 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.
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
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.
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.
As the rock rises, its horizontal component of velocity remains constant. This is because there is no horizontal force acting on the rock to change its velocity in that direction. The only force affecting the rock's motion is gravity, which acts vertically, causing its vertical component of velocity to decrease until the rock reaches its peak height and then falls back down.
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
It will increase until you hit 45 degrees and the it will begin to drop.
Horizontal position is strate 90 degree
For the projectile to land at the same distance with the same initial speed, it must be launched at an angle of 15 degrees from the horizontal. This is because the range of a projectile is maximized when launched at a 45-degree angle. So, launching at 15 degrees in the opposite direction of 75 degrees should bring the projectile to the same landing point.
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