Jan 13, 2010
15
0
0
Status
This is passage 5 from section I, translational motion.

A cannon is located on the edge of a cliff 100 m above a flat, sandy canyon floor. three cannon balls are launched from the cannon with the same initial speed of 50m/sec but at different angles. the cannon balls are identical except for their color. a red ball launched horizontally. white ball launched at 30degrees above horizontal. blue ball launched 60 degrees above horizontal.

If it were possible to launch all three cannon balls simultaneously, which order would they hit the ground?

i know you do not need to use quadratic equations for the mcat, but because i got the answer wrong i went ahead and did it anyway. i used the y = (vinitial)(t) + (1/2)(g)(t2) equation to solve for t. from the values of t that i got, i got that the red ball would strike the canyon floor at 4.5 seconds, the white ball would strike at around 2.7 seconds and the blue ball strike at 2.1 seconds. what am i doing wrong??
 
Mar 12, 2010
16
0
0
Status
Pre-Medical
Are you using the initial y velocity?

The only difference in the balls is their angle. This is a problem that doesn't require math. The ball with the lowest initial y velocity hits the ground first. The ball being launched horizontally has an angle of zero, giving it the lowest initial y velocity. The ball being launched at 30 degrees is second. The ball at 60 degrees hits last. So the answer is red, white, blue
 
Last edited:
Jan 13, 2010
15
0
0
Status
Are you using the initial y velocity?

The only difference in the balls is their angle. This is a problem that doesn't require math. The ball with the lowest initial y velocity hits the ground first. The ball being launched horizontally has an angle of zero, giving it the lowest initial y velocity. The ball being launched at 30 degrees is second. The ball at 60 degrees hits last. So the answer is red, white, blue

So let me show my calculations to be completely clear on where i'm going wrong:

for RED, i plug in the following:

y = (voy)(t) + (1/2)(g)(t2)
100 = 0 + (1/2)(10)(t2)
t = sqrt(20)

for WHITE, i plug in the following:

y = (voy)(t) + (1/2)(g)(t2)
100 = (50sin30)(t) + (1/2)(10)(t2)
t = around 2.6 sec

for WHITE, i plug in the following:

y = (voy)(t) + (1/2)(g)(t2)
100 = (50sin60)(t) + (1/2)(10)(t2)
t = around 2.1 sec

so i get blue, white and red...?? where am i going wrong??
 
Jun 14, 2009
800
3
0
Status
Pre-Dental
Well, if you're going to use +10m/s^2 for g, your initial velocities for the white and blue cannonballs need to be -50sin30 and -50sin60, respectively. +50sin30 and +50sin60 implies they were fired DOWNWARD at those angles.
 

unique135

10+ Year Member
Aug 13, 2008
146
1
0
Toronto
Status
Pre-Medical
Why use 100 for white and blue balls?

You will need to calculate the maximum height reached by each ball first.
 
Jan 13, 2010
15
0
0
Status
Well, if you're going to use +10m/s^2 for g, your initial velocities for the white and blue cannonballs need to be -50sin30 and -50sin60, respectively. +50sin30 and +50sin60 implies they were fired DOWNWARD at those angles.


GREAT. Thank you.

As for the height, I'm pretty sure you use y =100 for all three balls. Can someone confirm?
 
Jun 14, 2009
800
3
0
Status
Pre-Dental
Why use 100 for white and blue balls?

You will need to calculate the maximum height reached by each ball first.
no, that part is fine. Any difference due to firing angle is taken into account in the (voy)(t) term.
 
Dec 23, 2009
352
0
0
Australia or NYC or CT
Status
Pre-Medical
This is passage 5 from section I, translational motion.

A cannon is located on the edge of a cliff 100 m above a flat, sandy canyon floor. three cannon balls are launched from the cannon with the same initial speed of 50m/sec but at different angles. the cannon balls are identical except for their color. a red ball launched horizontally. white ball launched at 30degrees above horizontal. blue ball launched 60 degrees above horizontal.

If it were possible to launch all three cannon balls simultaneously, which order would they hit the ground?

i know you do not need to use quadratic equations for the mcat, but because i got the answer wrong i went ahead and did it anyway. i used the y = (vinitial)(t) + (1/2)(g)(t2) equation to solve for t. from the values of t that i got, i got that the red ball would strike the canyon floor at 4.5 seconds, the white ball would strike at around 2.7 seconds and the blue ball strike at 2.1 seconds. what am i doing wrong??
I don't think you even need to do any calculations for this problem.
The red ball has no initial y velocity so it must reach the ground first. then, the white ball will the hit ground because it has a little y more velocity initially. then, the blue ball will hit because it had even more y velocity initially.

also, you can check this, by writing the equation for motion in Y as
-100 = viy*sin(theta)*t - 5t^2

if we differentiate with respect to t, then we find that
viy*sin(theta) = 10*t
thus as theta goes up, t goes up.
 

Geekchick921

Achievement Unlocked: MD
Moderator Emeritus
10+ Year Member
Nov 6, 2007
8,829
165
281
34
Status
Resident [Any Field]
I don't think you even need to do any calculations for this problem.
The red ball has no initial y velocity so it must reach the ground first. then, the white ball will the hit ground because it has a little y more velocity initially. then, the blue ball will hit because it had even more y velocity initially.
This. Try not to waste time doing calculations you don't have to do, it'll save you a lot of time. :)