# Height of a projectile at two points?

• Vitani11
In summary, the homework statement states that a projectile is fired with a velocity v such that it passes through two points both at a distance h above the horizontal. If the gun is adjusted for maximum range, the separation of the points is given by d = (v/g)(√v2-4gh).
Vitani11

## Homework Statement

A projectile is fired with a velocity v such that it passes through two points both at a distance h above the horizontal. Show that if the gun is adjusted for maximum range, the separation of the points is

d = (v/g)(√v2-4gh)

## Homework Equations

y = -gt2/2 + vsinθt
x = vcosθt

for max range R = v2/g
θ = π/4

## The Attempt at a Solution

substituting theta into the two equations gives me
h = -gt2/2 + vt/√2

d = (v/√2)t

I replaced y with h to find the two times at which the ball is at h (both going up and down), then I used those times I found in the d equation to find the distance at both points at both times, and then took the difference between "d1" and "d2" where d2 is the point after max height and d1 is the point before. Is this not the right method? I get that t = h and t = stuff from the h equation... and t = h is obviously not true. Is that the issue?

Vitani11 said:
Is this not the right method?
Looks right.
Vitani11 said:
I get that t = h
Then you are making an algebraic error. You can use dimensional analysis to see where it goes wrong. If you cannot spot it, please post all your working.

Vitani11
h = -gt2/2 + vt/√2 to begin. I pull a t out and it becomes h = t(-gt/2+v/√2) so h = t and -gt/2+v/√2 = h

:/

Vitani11 said:
h = t(-gt/2+v/√2) so h = t
By what logic?

Vitani11
Vitani11 said:
h = -gt2/2 + vt/√2 to begin. I pull a t out and it becomes h = t(-gt/2+v/√2) so h = t and -gt/2+v/√2 = h

:/
That looks suspiciously like what you would do if you were trying to find the zeros of an equation. i.e. t(-gt/2+v/√2) = 0 so t = 0 or
(-gt/2+v/√2) =0. That isn't at all correct for h.

Vitani11
Finished. Thank you.

## 1. What is the formula for calculating the height of a projectile at two points?

The formula for calculating the height of a projectile at two points is h = vy^2 / 2g, where h is the height, vy is the vertical velocity, and g is the acceleration due to gravity.

## 2. How do you determine the vertical velocity of a projectile?

The vertical velocity of a projectile can be determined by using the formula vy = v0y + gt, where v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time elapsed.

## 3. What factors can affect the height of a projectile at two points?

The factors that can affect the height of a projectile at two points include the initial velocity, angle of launch, air resistance, and gravitational force.

## 4. Can the height of a projectile at two points be negative?

Yes, the height of a projectile at two points can be negative if the projectile is launched from a height and lands at a lower height, or if it is launched downward.

## 5. How does air resistance affect the height of a projectile at two points?

Air resistance can decrease the height of a projectile at two points by slowing down its vertical velocity and reducing the distance it can travel. It can also cause the projectile to deviate from its intended trajectory.

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