At which point in its path does a projectile have the least speed?

At which point in its path does a projectile have the least speed?

top
The minimum speed of a projectile occurs at the top of its path. If it is launched vertically, its speed at the top is zero. If it is projected at an angle, the vertical component of velocity is still zero at the top, leaving only the horizontal component.

What is least speed in a projectile?

Expert Answer:

• lowest speed of projectile motion is when the object reaches the maximum height.
• At the point of maximum height vertical component of velocity is zero and the particle has only horizontal component.
• Hence lowest speed of the object in projectile motion is ( 5 cos30 ) m/s , i.e. 5×√3/2 = 4.33 m/s.

At which point of the projectile does a body have smallest speed Why?

As at the highest point, there is only a horizontal component of the velocity present. Thus, at the highest point of the path, the projectile has the smallest speed.

How do you find the minimum speed of a projectile?

Separately, we know that, for a projectile for a given range R (on the same level), the minimum launch velocity is given by v∗=√gR. Assume that, for another given range r, the minimum launch velocity is w∗=√gr.

At what point or points in its path does a projectile have its maximum speed?

The speed is maximum at the highest point. At the highest point of the trajectory of a projectile, its speed is minimum. This is because the horizontal speed of the projectile remains constant.

At what point does a projectile have maximum speed?

Answer: In a Trajectory the projectile would have Minimum velocity at Trajectory point ( At the maximum height ) . & Possess Maximum velocity at initial projection point (due to initial acceleration) & at the point when projectile hits the ground .

At what point of the projectile path the speed is minimum at which point is it maximum?

highest point
The speed is maximum at the highest point. At the highest point of the trajectory of a projectile, its speed is minimum. This is because the horizontal speed of the projectile remains constant.

What is the speed of the projectile when it is at the highest point in its trajectory?

zero
At a projectile’s highest point, its velocity is zero.

At what point of the projectile path the speed of particle is minimum at which point speed is maximum?

it is minimum at the highest point and maximum when it is launched.

At what point projectile has maximum and minimum speed?

At what point of projectile motion is the acceleration and velocity perpendicular to each other?

At the top most point.

When a projectile is projected at a certain angle with the horizontal?

When a projectile is projected at a certain angle with the horizontal, its horizontal range is Rand time of flight is T1. When the same projectile is throwing with the same speed at some other angle with the horizontal, its horizontal range is R and time of flight is T2.

When does a projectile have the least speed?

Since only the vertical velocity changes, the lowest speed (i.e., the least magnitude of total velocity) must occur when the vertical component is zero — at the maximum height, just as the projectile stops rising and begins to fall.

How is the velocity of an object resolved?

Here is how! In the projectile motion,when object follows it’s trajectory.since projectile motion is a 2D Motion therefore it’s velocity can be resolved into two components I.e. vertical and horizontal velocity . Vertical velocity is dragged by the acceleration due to gravity.

Where does a ball have the highest velocity?

The ball has highest point where the vertical velocity is zero because until that point only the body is capable of moving up since velocity was in upward direction. Now it is Velocity has horizontal and vertical component when thrown making some angle with the ground.

How to calculate the range of a projectile?

The formula for the range of the projectile is Range = v^2 * sin 2 (theta) / g. I used g = 9.8 m/s^2 and a launched angle theta= 45 degrees. At 45 degrees sin 2 (theta) = 1. So with a range of 500 m, the equation is 500 = (v^2 * 1) / 9.8.