Q&A

How will you compare the range of a projectile fired at 30 and 60 angles?

How will you compare the range of a projectile fired at 30 and 60 angles?

The range is also the same for a 30° and a 60° launch angle. As a final example, the range is the same for a 40° and a 50° launch angle.

How do you find the maximum height of a horizontal projectile motion?

h = v 0 y 2 2 g . h = v 0 y 2 2 g . This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

At which angle will the projectile have a maximum height?

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

READ ALSO:   How many eggs should I eat before a workout?

What is maximum range of a projectile?

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

What is the maximum height of projectile thrown from the ground?

Maximum height of projectile thrown from ground is given by u 2 sin 2 θ 2 g and if the projectile is projected from a height H, then the maximum height attained by projectile during it’s flight is H + u 2 sin 2 θ 2 g as measured from the ground

What is the vertical component of velocity or speed at maximum height?

Now, what is the vertical component of velocity or speed in vertical direction at maximum height? Well, it is zero at that point i.e. v y = 0 at maximum height The horizontal range and maximum height of a projectile are equal – When?

How do you find the range of a projectile motion?

if α = 45°, then the equation may be written as: hmax = h + V₀² / (4 * g) and in that case, the range is maximal if launching from the ground (h = 0). if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that’s the case of horizontal projectile motion.

READ ALSO:   How many three digit numbers can be formed with the digits 1 2 5 4 6 The elements are repeated?

What is the maximum height of the object at the point?

Maximum height of the object is the highest vertical position along its trajectory. The object is flying upwards before reaching the highest point – and it’s falling after that point. It means that at the highest point of projectile motion, the vertical velocity is equal to 0 (Vy = 0). 0 = Vy – g * t = V₀ * sin(α) – g * th.