### Projectile_Motion_at_an_Angle

```Projectile Motion at an Angle
Katya Fatakhova
Lisa Levenson
Kaitlin Magee
Daniel Wittenberg
The Basics
• Figuring out the different parts of a projectile
launched at an angle require a few more steps
than when it is just launched horizontally
• There are more concepts to it and there are
different times that are used for different
equations.
• The best example would be to look at how a
ball is kicked (next slide)
-When the ball is first hit, it is at the fastest speed it can go, this
is called the instantaneous velocity.
-Once it leaves your foot, gravity acts on the vertical velocity
of the ball. This number will decrease as the ball gets
higher because gravity continues to pull down at it. Then it will
increase to its original value as the ball gets closer to the ground.
- When the ball is at its highest peak, the vertical velocity is
ZERO! This is because it is neither moving up nor down.
- The horizontal component of velocity stays the same all
throughout
SEE IT!
ure=related
VERTICAL PROJECTILE MOTION WITH DIFFERENT
COMPONENTS OF VELOCITY!
Figuring out the components
-When you have the instantaneous velocity, to find out what
the vertical component of velocity is, you use the
equation:
vy=vi(sinθ)
-When you have the instantaneous velocity, to find out what
the horizontal component of velocity is, you use the
equation:
vx=vi(cosθ)
-Since the ball makes an arc shaped movement
after it is kicked, it also has a maximum
height as well as a range, which is the total
distance that it traveled.
Figuring out the components Part II
-When finding the maximum height (dy or hmax),the equation
used is:
dy=vyt+(1/2)ayt2
vy is the vertical component of velocity and the ay is the
gravity that pushes down on the vy, which is (-9.81 m/s2)
As for the t (time) used in this equation, it is not the same
time as used in horizontal projectile motion (cont. on next
slide)
-Since the time it takes half to the time reach the maximum
height as it does to complete a projection, the term tup (time up)
is used. The equation for this t is:
t=(-vy)/(a)
Because of the parabolic curve of the motion, the ball is at its
highest point halfway through the complete cycle. Therefore, tup
is the time used to find dmax
dy-max = (1/2)(a)(tup)2
(When the object is at its highest point, its vertical velocity is zero.)
Range
-Now to find out the range, the equation used is:
dx=vxt
-The vx is the horizontal component of velocity and the t this
time is the total time, so you could use either the equation
ttotal = √(2d/a)
or you can just use the tup from before and multiply it by 2
(tup = ttotal/2)
*NOTE*
The maximum range of a projectile launched
at a given speed is achieved when the angle
with the horizontal is 45°
(cos 45º) = 1
Animation:
183. The maximum height reached by the ball is approximately
(1) 1.2 m (2) 2.5 m (3) 4.9 m (4) 8.5 m
183. The maximum height reached by the ball is approximately
(1) 1.2 m (2) 2.5 m (3) 4.9 m (4) 8.5 m
Explanation
-So you have to use the equation:
dy=vyt+(1/2)ayt2
-You plug in all of the values you know:
(the tup is the t given divided by two so it's 1/2)
dy=(4.9 m/s)(0.5 s)+(1/2)(-9.81 m/s2)(0.5 s)2
dy= 1.2 m
Projectile Motion at an Angle
An archer uses a bow to fire two similar arrows with the same
string force. One arrow is fired at an angle of 60 degrees with
the horizontal, and the other is fired at an angle of 45 degrees
with the horizontal. Compared to the arrow fired at 60 degrees,
the arrow fired at 45 degrees has a
(A) Longer flight time and shorter horizontal range
(B) Shorter flight time and longer horizontal range
(C) Shorter flight time and shorter horizontal range
(D) Longer flight time and longer horizontal range
Projectile Motion at an Angle
An archer uses a bow to fire two similar arrows with the same
string force. One arrow is fired at an angle of 60 degrees with
the horizontal, and the other is fired at an angle of 45 degrees
with the horizontal. Compared to the arrow fired at 60 degrees,
the arrow fired at 45 degrees has a
(A) Longer flight time and shorter horizontal range
(B) Shorter flight time and longer horizontal range
(C) Shorter flight time and shorter horizontal range
(D) Longer flight time and longer horizontal range
Explanation
dx = (vx)(time)+(1/2)at2
- There is no horizontal acceleration in this problem, so the
horizontal velocity will stay constant(neglecting air resistance)
So the equation becomes, dx = vx(t)
- To find vx, you have to multiply the original diagonal velocity
by cosθ
cos(45º) = 1- This means that a projectile fired at 45º will
always have the longest horizontal distance possible.
cos (60º) = 0.5
- If the two arrows were fired with the same initial velocity,
the one fired at 45º has a larger horizontal velocity, and
therefore travels farther horizontally.
Explanation (continued)
From the formula d=vi(t) + (1/2)at2 we find that
t = √(2dy/a) . Firing the arrow at 60º causes dy to be
increased, and therefore makes the flight time longer.
Summary
- The larger the angle, the longer the flight time.
- The closer the angle is to 45º, the longer the horizontal
distance
An airplane flying horizontally at 150m/s drops a bomb from a
height of 2500m. How far in front of the target must the bomb
be released?
This problem must be solved in two steps:
1) Find the time it takes for the bomb to hit the ground
dy = viy + (1/2)at2
(The initial y velocity is 0.)
2,500m = (1/2)(9.81m/s2)t2
2,500m = 4.9m/s2(t)2
2,500m/4.9m/s2 = t2
t = √(510s2)
t = 23s
HOMEWORK
Homework will be given out on a worksheet in class.
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