# Projectile: questioning need to double time

• Newlander
In summary, a ball is thrown straight upward with a velocity of 18 m/s and disregarding air resistance. Using the equation vf = vi + at, the time it takes for the ball to strike the ground can be calculated as t = [(0 m/s - 18 m/s)/-9.8 m/s2] = 1.8 s. This approach assumes that the ball is launched from ground height and reaches its maximum height before accelerating back down to the ground. The point of apogee, or highest height achieved, is the halfway point for the entire trajectory.
Newlander

## Homework Statement

"A ball is thrown straight upward with a velocity of 18 m/s. How much time passes before the ball strikes the ground? Disregard air resistance."

vi = 18 m/s
a = -9.8 m/s2
t = ?
[not sure: vf = 0 m/s]

vf = vi + at

## The Attempt at a Solution

vf = vi + at
t = [(vf - vi)/a]
t = [(0 m/s - 18 m/s)/-9.8 m/s2] = 1.8 s

Concerns:
--First, I understand that if I have final velocity as 0 m/s, that refers to when the ball has reached its maximum height--not its pre-impact final velocity.
--Second, with the first comment in mind, I'm not sure my approach is appropriate for this particular problem--and would appreciate guidance if it's not.
--Third and finally, some of my classmates insist that the correct answer is about double the answer I arrived at because "you just multiply the answer by 2 since it's only gone half of the distance" as I have conveyed above. I understand this thinking to some degree, but its simplicity seems to suggest the ball is projected from the ground--rather than simply thrown from what I'm gathering is knee-height. I would love to get your thoughts.

There does not appear to be any mention of knees, knee-height, or other initial launch heights in the problem statement, so it's best to assume that it's zero -- launched from ground height.

The ball goes up, decelerating all the way, achieves its maximum height (and zero velocity at that instant), then begins its downward journey, accelerating all the while, until it strikes the ground. By symmetry, it's velocity immediately before striking the ground will be -18m/s.

Your classmates are correct that the point of apogee (highest height achieved) is the halfway point for the entire trajectory.

Thanks very much for the prompt and informative reply!

## 1. What is a projectile?

A projectile is any object that is launched into the air and follows a curved path due to the influence of gravity.

## 2. How is the time of flight of a projectile calculated?

The time of flight of a projectile can be calculated using the formula t = 2 * v * sin(theta) / g, where t is the time of flight, v is the initial velocity, theta is the angle of launch, and g is the acceleration due to gravity.

## 3. Why is there a need to double the time of flight for a projectile?

Doubling the time of flight allows for a more accurate calculation of the distance traveled by the projectile. This is because the projectile follows a parabolic path and spends equal amounts of time in the upward and downward portions of its trajectory. Doubling the time accounts for both of these portions.

## 4. Is the initial velocity the only factor that affects the time of flight of a projectile?

No, the time of flight is also affected by the angle of launch and the acceleration due to gravity. A higher initial velocity or a lower angle of launch will result in a longer time of flight, while a higher acceleration due to gravity will result in a shorter time of flight.

## 5. Can the time of flight of a projectile be affected by air resistance?

Yes, air resistance can affect the time of flight of a projectile by slowing it down and reducing its overall distance traveled. In most calculations, air resistance is ignored, but it can be a significant factor in real-life scenarios.

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