At what angle horizontal range is equal to maximum height?

45∘
The angle of projection at which the horizontal range and maximum height of projectile are equal is. 45∘

What is the angle of projectile for which the range and maximum height become equal?

A projectile launched on level ground with an initial speed v0 at an angle θ above the horizontal will have the same range as a projectile launched with an initial speed v0 at 90° − θ and maximum range when θ = 45°.

How maximum horizontal range and maximum height are related for a projectile?

The maximum height of the projectile is when the projectile reaches zero vertical velocity. From this point the vertical component of the velocity vector will point downwards. The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object.

At which angle height of projectile is maximum?

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

What is the relation between range and maximum height if theta is 45?

The horizontal range is maximum when the angle of projection is 45°. That is, range is 4 times the maximum height attained by a projectile.

What angle gives the maximum range for a projectile?

45 degrees
A projectile, in other words, travels the farthest when it is launched at an angle of 45 degrees.

What is the maximum height of projectile with angle of projection?

If the angle of projection is 75.96 degrees the maximum height is equal to the horizontal range. At which angle height of projectile is maximum? This is not so much a math question as a physics question.

What angle is the maximum height equal to the horizontal range?

If the angle of projection is 75.96 degrees the maximum height is equal to the horizontal range. At what angle is the horizontal range equal to 8 times the maximum height in projectile motion?

What is the formula for the range of the projectile?

The formula for range is Range = v^2 sin2θ / g and for maximum height is dymax = (v sinθ)^2 / 2g. You may continue from here. You can cancel sinθ from both sides of the equation. The value of the angle θ is more I am giving you the formulas for solving the range of the projectile and its maximum height.

What is the horizontal component of the velocity at maximum height?

The horizontal component of the velocity does not change. At the maximum height the vertical velocity is equal to 0. As this is equal to the range: If the angle of projection is 75.96 degrees the maximum height is equal to the horizontal range.