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At what projection angle will the range of a projectile equal to 2 times its maximum height?
75.96... degrees from the horizontal.
Let the projectile be launched with speed v at an angle θ degrees above the horizontal. Then its vertical speed component is v sin θ, and its horizontal component is v cos θ.
The time of flight is t = 2 v sin θ / g, so that the range R is given by
R = v t cos θ = 2 v^2 sin θ cos θ / g.
The maximum height is given by H = (v sin θ)^2 / (2 g).
(You may think of this is as merely an application of the standard result that, dropping from rest in the second half of the flight, the downward speed u = v sin θ must be related to the downward acceleration a and distance d travelled by u^2 = 2ad. In this context, where a = g and d = H, that is equivalent to the conservation of kinetic plus potential energy, of course.)
If R = H, then v^2 sin^2 θ / (2 g) = 2 v^2 sin θ cos θ / g.
Therefore sin θ / cos θ = tan θ = 4.
Thus θ = arctan (4) = 75.96... degrees.
Live long and prosper.
EDIT : Note that the following answer contains a small error --- the formula presented for the maximum height H lacks the divisor 2 that it should properly have. Only with its incusion could one obtain the general result that
tan θ = 4H/R.
What else can I help you with?
What is the angle of projection if horizontal range is twice than height?
The angle of projection can be determined using the relationship between the horizontal range (R) and maximum height (H) of a projectile. If the horizontal range is twice the height, we can use the formula ( R = \frac{v_0^2 \sin(2\theta)}{g} ) and ( H = \frac{v_0^2 \sin^2(\theta)}{2g} ). Setting ( R = 2H ) leads to the conclusion that the angle of projection ( \theta ) is 60 degrees.
If you keep the velocity of projection constant and change the angle of projection from 75 to 45 what will happen to the horizontal distance the projectile travels?
The horizontal distance will be doubled.
What angle a projectile thrown to achieve maximum vertically distance traveled?
90
How does the maximum altitude of the projectile change as the launch angle is increased from 30 to 45 above the horizontal?
As the launch angle of a projectile increases from 30 to 45 degrees, the maximum altitude generally increases. This is because a higher launch angle allows for a greater vertical component of the initial velocity, which contributes to a higher peak in the projectile's trajectory. However, beyond 45 degrees, the altitude will begin to decrease as the horizontal component of the velocity becomes less efficient for achieving height. Thus, the maximum altitude reaches its peak at or around 45 degrees for a given initial velocity.
What is the formula for finding angles if a ball should be projected to attain the range and height when a ball is thrown upwards?
To determine the angles for projecting a ball to achieve a specific range and height, you can use the following formulas from projectile motion. The range ( R ) is given by ( R = \frac{v^2 \sin(2\theta)}{g} ), where ( v ) is the initial velocity, ( g ) is the acceleration due to gravity, and ( \theta ) is the launch angle. The maximum height ( H ) can be calculated using ( H = \frac{v^2 \sin^2(\theta)}{2g} ). By manipulating these equations, you can solve for the angle ( \theta ) based on the desired range and height.
The range of projectile is maximum when the angle of projection is?
The range of projectile is maximum when the angle of projection is 45 Degrees.
The angle of projection of a projectile for which the horizontal range and maximum height are equal?
45 degrees.
A projectile is fired in such a way that its horizontal range is equal to 14.5 times its maximum height What is the angle of projection?
15.42 degrees
What is the factor influence the distance and time in projectile motion?
projection speed projection angle projection height
How high does projectile go?
The maximum height of a projectile depends on its initial velocity and launch angle. In ideal conditions, the maximum height occurs when the launch angle is 45 degrees, reaching a height equal to half the maximum range of the projectile.
How does angle of projection affect the maximum height?
The angle of projection affects the maximum height by determining the vertical and horizontal components of the initial velocity. At 90 degrees (vertical), all the initial velocity is vertical which results in maximum height. As the angle decreases from 90 degrees, the vertical component decreases, leading to a lower maximum height.
What are the effect of changing the angle of projection of the magnitude of range maximum height and time of flight?
Changing the angle of projection affects the magnitude of range, maximum height, and time of flight. A higher angle will decrease the range and increase the maximum height while maintaining the time of flight. A lower angle will increase the range and decrease the maximum height while also maintaining the time of flight.
What is the relationship between range and height of projectile?
The range of a projectile is influenced by both the initial velocity and launch angle, while the height of the projectile is affected by the launch angle and initial height. Increasing the launch angle typically decreases the range but increases the maximum height of the projectile.
How does the launch angle relate to the initial speed of the projectile?
The launch angle and initial speed of a projectile are both factors that determine the range and height of the projectile. A higher launch angle with the same initial speed will typically result in a longer range but lower maximum height. Conversely, a lower launch angle with the same initial speed will result in a shorter range but a higher maximum height.
What is the angle of projection if horizontal range is twice than height?
The angle of projection can be determined using the relationship between the horizontal range (R) and maximum height (H) of a projectile. If the horizontal range is twice the height, we can use the formula ( R = \frac{v_0^2 \sin(2\theta)}{g} ) and ( H = \frac{v_0^2 \sin^2(\theta)}{2g} ). Setting ( R = 2H ) leads to the conclusion that the angle of projection ( \theta ) is 60 degrees.
How can one determine the maximum height reached in projectile motion?
To determine the maximum height reached in projectile motion, you can use the formula: textMaximum height left(fracv02 sin2(theta)2gright) where ( v0 ) is the initial velocity, ( theta ) is the launch angle, and ( g ) is the acceleration due to gravity. By plugging in these values, you can calculate the maximum height the projectile reaches.
What is the method for determining the angle of projection in projectile motion?
The angle of projection in projectile motion is determined by using the formula: arctan(vy / vx), where is the angle of projection, vy is the vertical component of the initial velocity, and vx is the horizontal component of the initial velocity.
