# Projectile motion (No initial velocity or time)

by PascalPanther
Tags: initial, motion, projectile, time, velocity
P: 23
This is the problem:
 You are an archer, you shoot up at a 45 degree angle above the horizontal, and hit a target at the same height as the horizontal 200m away. 1.Find the time the arrow spends in the air. 2. Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?
Let's ignore the second part, because I can't even get to there yet (sounds simple enough if I figured out the first part).

What I am stuck on, is the fact I have no initial velocity or time. So I have two unknowns.
Here is what I think I know. The acceleration is 9.8 m/s^2 due to gravity. Since I have the same y position, at the highest point P, where the velocity is 0, it should be exactly in the middle? So 100m.
If I use, 100m as x when y is at it's highest, I can turn that into a right angle. And do: (100)*tan 45 degrees = 100m. So the height at point P is 100m.

I'm assuming these are the only equations I am suppose to use:
v= v(i) + at
x = x(i) + v(i)t + 1/2at^2
v^2 = v(i)^2 + 2a(x-x(i))

x=100m
y=100m
t=?
v(i)=?
a=9.8 m/s^2

and also for y, since they are independent. And I think x and y have the same answer in this case, because they both have the same magnitude (at least cut in half). However, it doesn't look like I can solve for anything in any of the equations.

What should I be doing next?
 P: 14 Basically for question 1 you need to have separate the information into a vertical component and a horizontal component. Vertical: u=?, v=0, a=-9.8 and x=100, as you calculated. From here use equation v^2=(u^2)+2as, where u=initial velocity and s=displacement. From this you schould get u=44.3m/s. Now Horizontal: velocity is always constant horizontally so u=v and a=0. So use the formula s=ut+0.5a(t^2) and substitute values and transpose to get t. You should end up with t=4.51 secs.
 Sci Advisor PF Gold P: 1,544 In your example, wouldn't u=44.3 be the total initial velocity. Then he'd need to extract the x-component for the next step giving a final time in the 6 to 7 second range?
 P: 14 Projectile motion (No initial velocity or time) Too true tony, i forgot to breakdown the u=44.3 into a horizontal component. Once broken down it should equal 31.32m/s and this correct value for u leads to t=6.38m/s. Sorry guys

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