How can you determine the landing point of a horizontally launched projectile if you know the distance it will fall and the initial velocity?

Answer 1

Firing from a hill Firing a projectile from an elevated position increases its range. If you know the initial velocity, you should be able to use the usual formulas for displacement (distance) in the horizontal and vertical directions to determine the initial vertical position.1 When you say the initial velocity is known, I assume that includes magnitude and direction. Since velocity is a vector, you should be able to calculate the vertical and horizontal components.2 If you know the horizontal velocity and the horizontal displacement (distance traveled), you should be able to calculate the time in flight. Once you determine the time in flight, you should be able to use that value in the formula for vertical displacement to determine the initial vertical displacement. Hint: The vertical displacement of the projectile when it hits the ground is zero (assuming you have selected the origin -- the axes of the plane in which the projectile is moving -- properly). ----------- 1. d = d0 + V0t + [1/2]at2, where d0 is the initial displacement, v0 is the initial velocity, and a is acceleration. For motion in the vertical direction, a = -g. For motion in the horizontal direction, a = 0 (for projectile problems). 2. Vx = Vcos(theta); Vy = Vsin(theta), where theta is the angle of elevation. Maximum range is achieved when theta = 45 degrees. At that angle, Vx = Vy.

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PLEASE NOTE: formatting has been messed up in this so things that are supposed to be raised to a power, the number is not a superscript. This needs to be relooked at.

Maximum range is achieved when theta=45o only if the vertical displacement is zero (i.e. the projectile begins and ends at the same elevation). If launched from a certain height h, the angle for maximum range is given by

anglemax = 1/2 cos-1 [(gh)/(v^2 + gh)]

Returning to the problem, let h = launch height, R = horizontal distance from base of launch site to landing spot, V = launch speed, A = launch angle and T = time in air. The horizontal component of the launch velocity is constant since there is no acceleration in that direction. Therefore:

Vx = R/T

V cosA = R/T

Solving for T:

T = R/[V cosA]

Consider the vertical part of the problem. This solution is given for a projectile launched from an angle above the horizontal so that the initial vertical component of the velocity is positive when the acceleration due to gravity (g) is negative. Also assumed is that the launch position is above the landing position. Let the initial position be the origin.

d = do + viT + 1/2aT^2

-h = (V sinA)T - 1/2gT^2

Substituting for T:

(equation A) -h = [VR sinA ]/[V cosA] - [gR^2]/[2V^2 cos2A]

(equation B) h = -R [tanA] + [gR^2]/[2V^2cos2A]

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If you want to find the launch angle for a given height and launch speed that gives the maximum range, multiply both sides of equation A by 2cos2(theta) and rearrange to get:

(g/v^2)R^2 - (2sinAcosA)R - 2hcos2A=0

Using these trig identities:

2sinAcosA = sin2A

cos2A = 1/2 [1 + cos2A]

the equation becomes:

(g/v^2)R^2 - (sin2A)R - h[1+cos2A] = 0

Solving for R using the quadratic formula:

R(A) = v^2/(2g)[sin2A + (sin22A + (8gh/v^2)cos2A)1/2]

Find the derivative of

R(A): R'(A)=v^2/(2g)[2cos2A+1/2(sin22A+(8gh/v^2)cos2A)-1/2(4sin2Acos2A+(8gh/v^2)(-2sinAcosA))]

Set this equal to zero to find angle (A) for maximum range (R).

Rearrange and use some trig identities to get:

1/(cos22A) - v^2/(gh)(1/cos2A) - (1+v^2/(gh))=0

Use the quadratic formula to solve for 1/(cos2A):

1/(cos2A) = (1/2)[v^2/(gh) + (v^4/(g^2h^2)+4(1+v^2/(gh)))1/2] 1/(cos2A) = v^2/(gh) + 1

A = 1/2cos-1[(gh)/(v^2+gh)]

Answer 2

Variables:

t = time

vx = velocity in x direction

vy = velocity in y direction

v0x = initial velocity in x direction

v0y = initail velocity in y direction

v0 = initail velocity

a = acceleration

Numbers are in meters and seconds

The projectile was launched horizontally, so we know that at t = 0, vx = v0 and vy = 0. The speed in the x direction will not change during the flight; it is equal to the initial velocity.

The particle will accelerate in the y direction due to gravity. First find how long the particle takes to reach the ground, and then find out how far it traveled in the x direction during that time.

Let's go through an example problem. Say that the projectile was launched horizontally at a speed of 200 m/s and it will fall 20 meters. How far will it travel horizontally?

First we need to know how it takes the projectile to reach the ground. We can use the equation v2 = v02+2ay to find out how fast the projectile is going in the y direction when it hits the ground. a is 9.8, which is acceleration due to gravity. It is positive here because in this example we are taking the speed at the end to be positive. v0 is 0 because the projectile is not moving in the y direction at time t = 0:

v2 = 2(9.8)(20), so v = 19.80 m/s

We can then use the equation v = v0+at to find out how long it took the projectile to reach the ground (you could say that v and a are both negative, you will get the same result):

19.80 = 0 + (9.8)(t), so t = 2.02 s

Now we need to find out how far the projectile traveled in the x direction in 2.02 seconds. The speed in the x direction is 200 m/s, so multiply that by the time:

(200)(2.02) = 404 m

The projectile in this example traveled 404 meters horizontally.

Answer 3

Only the vertical component of motion determines how long a projectile is in the air. The equation to use is y = 1/2gt2, where y is vertical displacement in meters in a downward direction (negative), g is acceleration due to gravity, -9.8m/s2, and t is time in seconds. To solve this equation for time, do the following:

t2 = 2y/g

t = √(2y/g)

* Because both y and g are negative, when they are divided, the result is positive.