Question

projectile motion help:
a missile is launched horizontally at A with the speed of V0 = 188m;s The Horizontal distance travelled by the missile is d = 1200m
a. calculate the time of flight
b. calculate the value of y
c. calculate the velocity at point b

Answer 1

@mathmate @sweetburger @Jadeishere

Answer 2

@mww @Anaise @quickstudent

Answer 3

Let's start by writing equations to represent the flight motion with respect to time. This is important if this is part of physical mathematics. Physics may use already defined equations which are somewhat equivalent.

So we start by considering acceleration in horizontal and vertical directions at time t=0.
At t = 0, the horizontal acceleration x'' = 0 while vertical acceleration is y'' = -g where g is acceleration due to gravity. i.e. there is a downwards acting force on the projectile.

Integrate these with respect to t to find the corresponding velocity

\[\frac{ dx }{ dt } = \int\limits 0 dt = Ot +C = C\]
When t = 0, v(x) = dx/dt = 188 so C = 188 m/s
\[\frac{ dy }{ dt } = \int\limits -g dt = -g t +C\]
When t = 0, dy/dt = 0 (no movement in vertical direction, so C=0)

Then we integrate once more each to get the displacements with time,
\[x = \int\limits 188 dt = 188t + C\]
When t = 0, x = 0 so x = 188t
\[y = \int\limits -g t dt = -\frac{ 1 }{ 2 } g t^2 + C\]
When t = 0, y = h (height of projectile at launch site)
\[y = h-\frac{ 1 }{ 2 }g t^2\]

Answer 4

Time of flight is the time required for the projectile to touch the ground.
The projectile meets the ground when y = 0
So solve y = h -gt^2 = 0 for t to get the time for which this occurs

Value of y (really h). You will need to use the range of flight of d=1200m to help you.
The time it lands is the same as the time that projectile flies 1200m since that distance was covered in landing.
So solve for t, x = 1200 = 188t so t = 1200/188
Now sub in this t into y to find out the distance dropped in that time
y=0 = h - 1/2gt^2 (sub in that t here and solve for h)

Velocity at b
Find dx/dt and dy/dt at b, the landing site by subbing in t = 1200/188
Then use \[v = \sqrt{(\frac{ dx }{ dt })^2+(\frac{ dy }{ dt })^2}\]

Answer 5

If the course was done in physics they may have taught you different equations instead:

\[v_y = u +at\]
\[s_y = ut +\frac{ 1 }{ 2 }at^2\]
\[v_x = u_x\]
\[s_x = u_xt\]

Similar principles apply to how to find the values set.

Answer 6

I've enclosed a powerpoint slide show giving my version of things. I'd add a few points to it. If the missile were dropped dropped from an aircraft, as in a bomber aircraft, then the bomber would have to release the bomb from the craft before the plane was over the target.
Relative to the aircraft, the bomb/missile would be seen to fall directly below them since each would have the same horizontal speed. Air resistance would be assumed to affect each of them the same, and so cancel itself out.
Relative to some hapless observer on the ground, the aircraft would be seen flying in a straight line, whilst the missile would be seen following a PARABOLA, though the terrified observer may not be clued up on the (un?)niceties of ballistics.
Since the speeds involved are very very small by speed of light standards, what I'm trying to get at above is that this CAN be seen as an example of LOW SPEED or Galilean relativity.
To get into high speed relativity (the Einstein - Mileva and Albert - stuff) needs speeds comparable to 300, 000, 000 m/s (speed of light in a vacuum). By way of comparison, I sonic jet aircraft is doing 330m/s, and, I think, a Mach2 supersonic jet is doing 660m/s. I forget the speeds that the ill fated UK/France collaboration of "Concorde" did.
I hope, obviously, I've got most of this right !
http://www.perendis.webs.com