Is my reasoning and method for obtaining this projectile motion question wrong?

[DrDreams]

Please let me know. I thought it was a better way than Kaplan did it. Speed beats precision right?

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[kraskadva]

Your links are broken

 

[DrDreams]

Your links are broken

links fixed. Thanks

 

[Cawolf]

I am not sure I agree with how you found the height at which the object explodes.

F = ma

a = F/m = (1.2 N)/(0.1kg) = 12 m/s^2 - g = 2 m/s^2

Height = 0.5at^2 = (0.5)(2 m/s^2)(5s)^2 = (1 m/s^2)(25 s^2) = 25 m


For the leftward piece why did you multiply the velocity by 0.71 if all it's velocity is horizontal?

I would find the height of explosion and then look at each piece, sort of how you did.

Left piece:

Time to Fall: y = y0 + vy0t + 0.5at^2

0 = 25 m + 0.5gt^2

t^2 = 2(-25 m)/(-10 m/s^2)

t = sqrt (5 s^2) = about 2.24 s

Distance traveled: x = x0 + v0xt

x = (-15 m/s)(2.24s) = about 34 meters left

Right Piece:

Time to Fall: y = y0 + vy0t + 0.5at^2

0 = 25m + v0sin(53.1)t + 0.5gt^2 (sin 53.1 somewhere between 0.71 and 0.87 approx 0.8)

0 = 25m + (25 m/s)(0.8)t - (5 m/s^2)t^2

t = [-b +/- sqrt(b^2 - 4ac)]/2a = [-20 +/- sqrt(20^2 - 4*(-5)*25)]/2*-5 = [-20 +/- sqrt(400 + 500)]/-10 = (-20 +/- 30)/-10

t = 5 s

Distance Traveled: x = x0 + v0xt

x = (25 m/s)(cos 53.1)(8s) (cos 53.1 between 0.5 and 0.71 about 0.6)

x = (25 m/s)(0.6)(5 s) = 75 m

So distance between them = 34m + 75 m = 108m

 

[Odi]

Your mistake starts in step 2 and continues in step 3.

Kaplans method of dealing with projectile motion involves too much formula math and not enough conceptual reasoning. TBR makes it very conceptual where all you need to use is x=1/2at^2 which simplifies to x=5t^2 since acceleration is usually gravity for any projectile motion or free fall problem.

There are two forces acting on the rocket: upward 1.2N and downward mg which equals 1N.

The net force:
1N - 0.2N = ma
0.2N = ma
a=2

How high does it go in 5 seconds?

y=1/2at^2
y=1/2(2)(5^2)
y=25m.

At 25 m the object explodes into two pieces.

Now we have a simple projectile problem where acceleration is gravity. Anytime you have a projectile problem like this, find the time in flight and use the simplified blue equation because that will solve everything.

Left piece flight time:
y= 5t^2
25=5t^2
t=2.2s

The object has a constant horizontal velocity of 15m/s to the left and travels for 2.2 seconds, how far will it go?
33m

Right piece flight time:

It has an upward velocity of 20m/s according to 25sin53.1.

Think of this conceptually, how long does it take gravity (-10m/s^2) to "eat" up an upward velocity of 20m/s? 2 seconds!

How high does the object travel?
y=5t^2
y=5(2)^2
y=20m

At the apex, the object is now 45m high (25m + 20m). How long does it take to come down to the ground from its apex?

y=5t^2
45=5t^2
t=3seconds.

The right side object is in motion for a total of 5 seconds (2 seconds up + 3 seconds down)

How far does it move in 5 seconds with a constant horizontal velocity of 15m/s? 75m

 

[DrDreams]

The correct answer was 115. Rather than going through all the steps I used 5 seconds as my time. Then used the angle in each direction to l find the x component of velocity. Then I just added them together.

Then I used the range formula of r=voxt in each direction.

So I can't use this reasoning going forward? I did use tbr's approach.