Homework Help: Throwing an object over the fence

1. Jul 11, 2011

forrestuv

1. The problem statement, all variables and given/known data
I need to throw an Object over a fence at a destinationPoint reached in a given time

startPointX=0
startPointY=0.4
destpointX=25
destpointY=0
time=3
fence height=1
fence posX = 10

3. The attempt at a solution
I decided to split up the problem in two components X,Y

initialSpeedX=(destPointX-startPointX) / time

giving the formula of a falling body:
distance = 1/2 × gravity × time^2 + intialspeedY × time

initialSpeedY=((destPointy - startPointy) - (0.5 * gravity * (time* time))) / time

now if I throw the object at an initial velocity of (initialSpeedX,initialSpeedY) it will hit the destpoint in the given time.

now I can calculate if the object at a certain time will hit the fence

timeToReachTheNet = (fencePosX-startPointX) / ((destPointX-startPointx) / Time)
fenceHitHeight = startPointy+( (gravity * (timeToReachTheNet^2) + 2 * initialSpeedY* timeToReachTheNet ) / 2);

if the fenceHitHeight is less than 1 it will hit the fence.

now, if the object hit the fence, how can I find the initialSpeedY so it will reach at least 1mt of height and cross the fence?

it seems that a simple equation is not enough... uhm..

maybe I could find the right equation to directly without the fence hit test..but this is the only method I know :(

2. Jul 11, 2011

rl.bhat

You have found initial speed x and initial speed y.
Now tanθ = (initial speed y) / (initial speed y)
Using kinematic equation, you can see that
yf = yi + x*tanθ - 1/2*g*x^2/(v^2*co^2θ)

= yi + x*tan(θ) - 1/2*g*x^2*sec^2(θ)/(v^2)
Put sec^2(θ) = 1 + tan^2(θ) and solve the quadratics to find tanθ.

3. Jul 11, 2011

forrestuv

really hard for me...I'm just a programmer :(
could you be go into details a little more?

would it require much time for you to write the final formula to calculate the initial velocity to make the object reach a destination point, crossing a fence?

I'm not able to do all the calculations required :(