# Thread: Physic - Velocity, distances, acceleration, and time

1. ## Physic - Velocity, distances, acceleration, and time

here's 3 problems i need to solve.. but i got all of them wrong and I don't understand why.. I believe i had the right formula/procedure. can someone show me how to do it? thanks

1. A football is thrown directly toward a receiver with an initial speed of 15.0 m/s at an angle of 23° above the horizontal. At that instant, the receiver is 18 m from the quarterback. In what direction and with what constant speed should the receiver run to catch the football at the level at which it was thrown?

2. A rocket is launched at an angle of = 45° above the horizontal with an initial speed vi = 57 m/s, as shown below. It moves for 25 s along its initial line of motion wth an acceleration of 22.2 m/s2. At this time, its engines fail and the rocket proceeds to move as a free body.
(a) What is the rocket's maximum altitude?
(b) What is the rocket's total time of flight?
(c) What is the rocket's horizontal range?

3. A science student riding on a flatcar of a train moving at a constant speed of 12.8 m/s throws a ball toward the caboose along a path that the student judges as making an initial angle of 65° to the horizontal. The teacher, who is standing on the ground nearby, observes the ball rising vertically. How high does the ball rise?

thanks so much for the help!!

2. Originally Posted by cbbplanet
1. A football is thrown directly toward a receiver with an initial speed of 15.0 m/s at an angle of 23° above the horizontal. At that instant, the receiver is 18 m from the quarterback. In what direction and with what constant speed should the receiver run to catch the football at the level at which it was thrown?

Here's the basic plan of attack for this kind of problem.

Always start by setting up and origin and coordinate system. I would put the origin where the ball was thrown with a +x axis in the direction of the horizontal component of the throw and a +y axis directly upward.

I will assume the ball is caught at the same height it was thrown at, since we have been given no information to the contrary.

Now list all of your information:
$x_0 = 0~m$ and $y_0 = 0~m$

$v_{0x} = 15~cos(23)$ and $v_{0y} = 15~sin(23)$

$a_x = 0~m/s^2$ and $a_y = -9.8~m/s^2$

The ball is caught at time t at $x = 18~m$ and $y = 0~m$.

Now, in this time, the receiver runs from the origin to x = 18 m in this same time t at a constant speed. So
$x = x_0 + vt$

$18 = vt$

$v = \frac{18}{t}$

So we need an expression for t.

Look back at the football data. We have essentially 5 equations at our disposal:
$x = x_0 + v_{0x}t$ <-- Since $a_x = 0~m/s^2$ this equation contains all the Physics.

$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

$y = y_0 + \frac{1}{2}(v_{0y} + v_y)t$

$v_y = v_{0y} + a_yt$

$v_y^2 = v_{0y}^2 + 2a_y(y - y_0)$

From the x equation we get
$18 = 15~cos(23)~t$

Or we could use the first y equation:
$0 = 15~sin(23)t - 4.9t^2$

My difficulty with this problem is that the two equations do not agree with what the time is. The only way I can correct for this is to assume that either the receiver is not at the level of the origin, or the quarterback isn't. If this is the case then we must choose the x equation to give us the time since this equation is not altered by a height change. I do not like the necessity of making this kind of logic chain, and I certainly would not expect a typical student to come up with it. It is standard in "throwing" problems like these to make the assumption that the projectile starts and ends at the same height unless otherwise explicitly stated in the problem.

So anyway I get that
$t = \frac{18}{15~cos(23)} = 1.30363~s$

Thus the receiver needs to run at
$v = \frac{18~m}{1.30363~s} = 13.8076~m/s$

The other problems can be done using a similar setup.

-Dan

3. ^ i got the exact same thing T = 1.2 s and the V = 13.8 m/s

the thing is it's telling me its wrong. so i'm not sure what's wrong. may be it is right and my teacher just entered the answer wrong.. for 3 questions straight? hmm

also another thing is that the receiver suppose to wrong TOWARD the quarterback to catch the ball because it will land at 16.6 m.. i'm not sure if it will effect anything much. but thanks so much for the help!! at least i know i'm on the right track

4. Originally Posted by cbbplanet
^ i got the exact same thing T = 1.2 s and the V = 13.8 m/s

the thing is it's telling me its wrong. so i'm not sure what's wrong. may be it is right and my teacher just entered the answer wrong.. for 3 questions straight? hmm

also another thing is that the receiver suppose to wrong TOWARD the quarterback to catch the ball because it will land at 16.6 m.. i'm not sure if it will effect anything much. but thanks so much for the help!! at least i know i'm on the right track
(sigh) Well that changes things. So for the football data $x = 16.6~m$ and for the receiver he has to start at 18 m?? This needs to be confirmed before a solution can be given.

-Dan

5. Yes, I caculated that the football's x value is 16.6 (but it wasn't given) the 18m was given in the problem though

6. 1. A football is thrown directly toward a receiver with an initial speed of 15.0 m/s at an angle of 23° above the horizontal. At that instant, the receiver is 18 m from the quarterback. In what direction and with what constant speed should the receiver run to catch the football at the level at which it was thrown?

The football will be thrown a maximum horizontal distance of
H = (15)cos(23deg)*T meters.

What is T?

At maximum heigth of the ball's path, the net vertical velocity is zero, so,
(15)sin(23deg) +(-10)t = 0 <----we assume g = 9.8 m/sec/sec.
t = 15sin(23deg) /9.8= 0.598 sec.
Meaning, the ball spent 0.586 seconds to reach the maximum height.

To reach H, the ball will spend the same 0.586 seconds again, so,
T = 2(0.598) = 1.196 sec.

Hence, H = 15cos(23)*1.196 = 16.514 meters.

The receiver is intially at 18 meters from the thrower, so the receiver must run towards the thrower to cover that
18 -16.514 = 1.486 meters...(the ball is thrown short)
So,
Let s = constant speed of the receiver,

distance = speed*time
[time is 1.196 sec]
1.486 = s(1.196)
s = 1.486 /1.196 = 1.242 m/sec

Therefore, the receiver should run towards the point of throwing, at 1.242 m/sec to catch the football. --------answer.

7. thanks i realized that i got the problem wrong only because of significant figure T_T i put 1.2 instead of 1.24.

anyways what i really need help is with number 2 or 3. i really am confused especially with them. whichever problem is easier for you, can you do them?

8. *2. A rocket is launched at an angle of = 45° above the horizontal with an initial speed vi = 57 m/s, as shown below. It moves for 25 s along its initial line of motion wth an acceleration of 22.2 m/s2. At this time, its engines fail and the rocket proceeds to move as a free body.
(a) What is the rocket's maximum altitude?
(b) What is the rocket's total time of flight?
(c) What is the rocket's horizontal range?

It moves for 25 s along its initial line of motion wth an acceleration of 22.2 m/s2.
I assume that means for 25 seconds, the rocket flew a straight line that is at 45 degrees above the horizontal, and that its net acceleration (including the effect of gravity) is 22.2 m/sec/sec in a straight line.

So, at the end of 25 seconds,
V(t) = Vo +at
V(25) = 57 +22.2(25) = 612 m/sec at 45 degrees above horizontal

After that, the rocket moves as a free body. Meaning, no more booster.

--------------------------------------------------------
(a) What is the rocket's maximum altitude?

From 0 to 25 seconds,
s = ([(Vo +V(25)]/2)*25 = ((57 +612)/2)(25) = 8362.5 meters at 45 degrees above the ground.
So,
h1 = (8362.5)sin(45deg) = 5913.2 meters vertically above the ground.

From just after 25 seconds to time the rocket reaches its maximum height,
h2 = 612sin(45deg)*t -(1/2)(9.8)t^2 ------------(i)

What is t?

At maximum height, vertical velocity is zero, so,
612sin(45deg) -(9.8)t = 0
t = 612sin(45deg) /9.8 = 44.158 sec.

So,
h2 = 612sin(45deg)*(44.158) -4.9(44.158)^2 = 9554.7 meters above h1.

Therefore, the rocket's maximum altitude is 5913.2 +9554.7 = 15,468 meters above ground. [That is almost 15 and a half kilometers above ground.] -------------answer.

------------------------------------------
(b) What is the rocket's total time of flight?

The total time until the rocket hits the ground?

With booster, 25 sec.
From h1 to h1 again, 2(44.158) = 88.316 sec.

From h1, on the flight downwards, to the ground,
-------Vertical velocity is the opposite of the vertical velocity at the h1 while on the way up, so, V = -612sin(45deg) = -432.75 m/sec
------h = -h1 = -5913.2 m
So,
s = Vo*t -(1/2)g*t^2
-5913.2 = -432.75*t -4.9t^2
4.9t^2 +432.75t -5913.2 = 0
Divide both sides by 4.9,
t^2 +88.32t -1206.78 = 0
t = {-88.32 +,-sqrt[(88.32)^2 -4(1)(-1206.78)]} /2(1)
t = {-88.32 +,-112.37}/2
t = -100 sec, or 12.025 sec
So,
t = 12.025 sec <------------from h1 to ground.

Therefore, total flight time = 25 +88.316 +12.025 = 125.34 seconds. ----answer.

---------------------------------------------------
(c) What is the rocket's horizontal range?

From 0 to 25 seconds,
x1 = h1 = 5913.2 meters

From just after 25 sec to (25 +88.316 = 113.316) sec,
x2 = 612cos(45deg)*(88.316) = 38,218.7 meters

From just after 113.316 sec to the rocket hitting the ground,
x3 = 612cos(45deg)*(12.025) = 5203.8 meters

Therefore, the rocket's horizontal range is 5913.2 +38,218.7 +5203.8 = 49,335.7 meters. [About 49 and 1/3 kilometers.] -----------answer.

--------------------------------------------------
If the answers above are wrong with your teacher, then your teacher considered the effect of gravity at the first 25 seconds of the flight.

We can solve for that too.

9. 3. A science student riding on a flatcar of a train moving at a constant speed of 12.8 m/s throws a ball toward the caboose along a path that the student judges as making an initial angle of 65° to the horizontal. The teacher, who is standing on the ground nearby, observes the ball rising vertically. How high does the ball rise?

The initial velocity of the ball is not known?

That means the throw is against the direction of the caboose.
So, the horizontal component of the initial velocity of the ball is equal to 12.8 m/sec since the flight of the ball is vertical---no horizontal displacement.

So,
(Vo)cos(65deg) = 12.8
Vo = 12.8 /cos(65deg) = 30.29 m/sec

And so, the vertical component of the initial velocity of the ball is
(Vo)sin(65deg) = 30.29sin(65deg) = 27.45 m/sec.

Hence,
max height, H = 27.45(t) -4.9(t^2)

What is t?

At H, vertical velocity is zero, so,
27.45 -9.8t = 0
t = 27.45/9.8 = 2.80 sec.

Therefore, H = 27.45(2.8) -4.9(2.8)^2 = 38.44 meters above the point of throwing. -------------answer.

10. wow, the answers are perfectly right! i have to figure what i did wrong.

what i did was break down the problem into 3 stages.

1st stage = the point from the ground to the point where the engine stopped working
2nd stage = the point where the engine stopped working to the highest altitude
3rd stage = the highest altitude to the point it landed back to Earth.

For each step I use the coordinate system in x direction and y direction and find out my missing value (ex. final velocity, final delta x)

THANKS SO MUCH FOR YOU GUYS HELP!! it really helped me!