velocity of 20 m/s, assume g=10 m/s2. Calculate: a) The time it will stay in the air? b) How far will the stone travel before it hits the ground (the range)? c) What will be the maximum height the stone will reach?
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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
A velocity time graph is still a velocity time graph - no matter the degree of detail that you look at it.
Ask Sir JB.
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Suppose a projectile is fired from a gun, we know that "g" remains constant and as we use horizontal component of velocity in range sov0 also remains constant. Only sin2θ responsible for change in range. The range will be maximum if sin2θ has its maximum value that is 1.for maximum range:sin2θ = 12θ = sin-1 (1)θ = 90/2θ = 45 (degree)therefor if projectile is projected with the angle of 45(degree) its range will be maximum.
body is projected with a velocity 3o m/s at an angle 30 degree with vertical find maximum height time of flight and range
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.
The degree is equal to the maximum number of times the graph can cross a horizontal line.
At the highest point of the particle's trajectory, its kinetic energy will be zero because it momentarily comes to a stop at that point. Potential energy will be at a maximum at this point.
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.
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.
The induced current is maximum when the axis of the conductor, its velocity, and the magnetic field lines are all mutually perpendicular.
w2hen the angle of projection is larger than 45 degree the height attained will be more but the range is again less.in this case angle is 30 degree which is less than 45 degree and 60 degree angle is greater than angle of 45 degree which have less range and is equal to range of 30 degree .so ranges are equal.
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.
The time the ball is in the air can be found using the vertical motion equation: time = 2 * (initial vertical velocity) / (gravity). The range can be calculated using the horizontal motion equation: range = (initial velocity)^2 * sin(2*launch angle) / gravity. The maximum height can be determined by finding the vertical component of the flight time and substituting that into the vertical motion equation: max height = (initial vertical velocity)^2 / (2 * gravity).
The weight and shape of the bullet, the speed at which the bullet leaves the gun, and the degree of tilt of the barrel above the horizontal.