Finding Projectiles' Flight Times....

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Two projectiles are launched with initial speeds of 100 meters per second, one straight up and the other at a 30-degree angle. The projectile launched straight up takes 20 seconds to return to the ground, while the one at a 30-degree angle takes 10 seconds. The calculations involve using the equation y = vt - 1/2gt^2, with adjustments for vertical velocity components. The vertical component for the angled projectile is calculated as 50 m/s, leading to its shorter flight time. The discussion highlights the importance of correctly applying physics equations to determine projectile flight times accurately.
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Homework Statement


Two projectiles are launched from the ground with initial speeds of 100 meters per second. One is launched straight up, the other at a 30-degree angle. How much time elapses between each's return to ground level?

Theta = 30
g = 10 m/s
Vinitial = 100

Homework Equations


y = vt - 1/2gt^2

The Attempt at a Solution


As you can see in the image, I attempted to use the equations that I THINK were correct for this problem. My professor just threw a bunch of equations up on the board, some of which he said we don't need for now and they're just there to be there...so, hopefully I did these right.

BLQfree.jpg
 
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The attachment doesn't make your approach clear. How did you go about solving this?

And, what answer did you get?
 
PeroK said:
The attachment doesn't make your approach clear. How did you go about solving this?

And, what answer did you get?
I wrote the formula that I used on the paper...it's right underneath the given variables. My answers were that the object thrown straight up in the air took 20 seconds to fall, and the object that was thrown at a 30 degree angle took 5 seconds to fall.
 
EthanVandals said:
I wrote the formula that I used on the paper...it's right underneath the given variables. My answers were that the object thrown straight up in the air took 20 seconds to fall, and the object that was thrown at a 30 degree angle took 5 seconds to fall.
Why 5 seconds?
 
PeroK said:
Why 5 seconds?
Because, after plugging in 0 for y (when the object hits the ground), I got the equation 0 = 50t-10t^2, and after solving for t, I got 5.
 
EthanVandals said:
Because, after plugging in 0 for y (when the object hits the ground), I got the equation 0 = 50t-10t^2, and after solving for t, I got 5.

So, why 20 seconds in the first case?
 
PeroK said:
So, why 20 seconds in the first case?

Don't you think the difference between 20 and 5 is too large?

Can you spot your mistake?
 
You can view from vertical perspective.. from ground to max for each projectile.

First, vy=100.. using vt=v-gt
t=v/g=100/10=10s

Second, vy=100.sin30=50 m/s
t=5seconds

Total time for projectile 1 is 20s
Time for projectile 2 is 10s
So difference/time elapse is 20-10=10s
 
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