Finding Projectiles' Flight Times....

In summary, two projectiles with initial speeds of 100 meters per second were launched from the ground, one straight up and the other at a 30-degree angle. Using the equations y = vt - 1/2gt^2, it was determined that the object thrown straight up took 20 seconds to fall and the object thrown at a 30-degree angle took 5 seconds to fall. The difference in time between their returns to ground level is 10 seconds.
  • #1
EthanVandals
55
2

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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  • #2
The attachment doesn't make your approach clear. How did you go about solving this?

And, what answer did you get?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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?
 
  • #8
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
 

1. What is a projectile?

A projectile is any object that is launched or thrown into the air and follows a curved path due to the force of gravity.

2. How do you calculate a projectile's flight time?

The flight time of a projectile can be calculated using the formula t = 2v*sin(theta)/g, where t is the flight time, v is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity.

3. What factors can affect a projectile's flight time?

The factors that can affect a projectile's flight time include initial velocity, launch angle, air resistance, and the height of the launch point relative to the landing point.

4. How can you use the flight time of a projectile in real-life situations?

The flight time of a projectile can be used to predict the trajectory and landing point of objects such as projectiles launched from a catapult, or the flight of a baseball hit by a player.

5. Can the flight time of a projectile be affected by external forces?

Yes, the flight time of a projectile can be affected by external forces such as wind, air resistance, or the rotation of the Earth.

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