Given height, and velocity thrown upward, find air-time

[bigman8424]

got a fun problem, but having some problem finishing it, a water balloon is thrown at 12 m/s from roof of building which has height of 30.0 m. what is time of flight, and vel. when hits ground.

im thinking the formula y = y(i) + v(i)sint + 1/2gt2 is used here.
y = 0 + 12*sin(0)t + 1/2(-9.8)t2
y = 0 + 0 - 4.9t2

not doing this problem right, someone please help me out

thx

 

[whozum]

The angle made with the ground is 90 degrees, not 0. Also, there is a change in final position, and you need to take it into account (hint: 30 should show up in your equation somewhere).

 

[bigman8424]

awww, got you, it's coming together now, the new formula i got is:
y = 30 + 12sin(90)t +1/2(-9.8)t2

solve for t?

 

[wisredz]

g=9.8 not -9.8. time of flight is 1.536275359. velocity at the time of impact would be 12+9.8*1.536275359=27,0554985182. Did I do anything wrong?

btw, I assumed that the balloon was thrown downwards. If it were thrown upward, velocity would be the same, flight of time would increase. but if we throw horizontally then, flying time would be 2,5 sec. velocity would be (24,5^2+12^2)^0.5=27,280945731407479928629640660381.

Anything wrong with any of the calculations I made?

 

[whozum]

g=9.8 not -9.8. time of flight is 1.536275359. velocity at the time of impact would be 12+9.8*1.536275359=27,0554985182. Did I do anything wrong?

btw, I assumed that the balloon was thrown downwards. If it were thrown upward, velocity would be the same, flight of time would increase. but if we throw horizontally then, flying time would be 2,5 sec. velocity would be (24,5^2+12^2)^0.5=27,280945731407479928629640660381.

Anything wrong with any of the calculations I made?

Yeah, they're useless to the OP because no one has any idea how you didi t.

 

[whozum]

awww, got you, it's coming together now, the new formula i got is:
y = 30 + 12sin(90)t +1/2(-9.8)t2

solve for t?

Yes that will give you the correct solution. Apply the quadratic formula, and remember to take the positive root.

 

[wisredz]

what I did is just some simple calculations. I assume you throw it horizontally, I also assume that there's no air friction...
Because there's no air friction the balloon's horizontal velocity will always be the same. It's clear that it will fall 30 m before hitting the ground and that it has no vertical speed, so finding t from the equation 30=0.5*9.8*t^2 would be the time the balloon has been airborn. the value's actually 2,47... but I took it 2.5. Anyway by this time the balloon will have gained a vertical speed of 2.5*9.8=24,5. So when it hits the ground it has a vertical speed of 24.5 and a horizontal speed of 12. applying the phythagoras theorem gives us the velocity at the moment of impact.
Is it clear now?

 

[whozum]

what I did is just some simple calculations. I assume you throw it horizontally, I also assume that there's no air friction...
Because there's no air friction the balloon's horizontal velocity will always be the same. It's clear that it will fall 30 m before hitting the ground and that it has no vertical speed, so finding t from the equation 30=0.5*9.8*t^2 would be the time the balloon has been airborn. the value's actually 2,47... but I took it 2.5. Anyway by this time the balloon will have gained a vertical speed of 2.5*9.8=24,5. So when it hits the ground it has a vertical speed of 24.5 and a horizontal speed of 12. applying the phythagoras theorem gives us the velocity at the moment of impact.
Is it clear now?

The balloon is thrown upwards with a velocity of 12m/s. Please read the problem. Also, the point of the forum isn't for you to solve the problem, its to help the poster do it.

 

[wisredz]

Then there are 2 parts to solve this problem. first part is about how much time is needed for the balloon to come back to the original height of 30 m. For a brief time The balloon will decelerate until it reaches it's highest point and there its velocity will be 0, and then it will start to acelerate downwards. The balloon's motion until it comes back to the initial height of 30 m can be studied in two parts: before it reaches the peak point(it decelerates in this part) and until it comes back to 30 m(It accelerates in this part). The time that passes for the balloon to reach the peak point and to drop again to 30 m is the same, that's why I study the motion in 2. Okay, now think of the first part like this. How much time would be needed for a particle thrown upward with the initial speed of 12 to reach the highest point of its motion? We know that at the highest point the velocity should be zero. Then we got this equation: 12-9.8t=0. t = 1,22 sec. remember that this the first part of its motion. Now it will drop in the same time to the level of the building's top. Its speed then will be 12 again because if you decelerate from 12 to 0 in 1,22 sec, then you will accelerate from 0 to 12 in 1,22 sec as well. So the first part of the problem is solved, it comes back to its initial height in 2.44 sec. Then we proceed to the other part. Now the problem is like this, a balloon is thrown downwards with the initial speed of 12m/sec, bla bla. Note that the only difference is the extra time we calculated earlier. It's clear that the balloon wil travel 30 m before hitting the ground. so 30=12t+0.5*9.8*t^2. the time needed for the balloon to hit the ground is found from this equation and is 1.53627. So the total airtime would be 1.53627+2,44=3.98 which is close to 4. Okay, now we know how long the balloon has been airborne before hitting the ground. As to its velocity then, all we need to do is calculate how much it would accelerate in 1,53627 sec. and sum this with the initial speed of 12. It would gain a speed close to 15 m/sec in 1.53627 sec. and we still have the initial speed of 12. 12+15=27 m/sec is the final speed.

Our answers are velocity=27m/sec. and time=4 sec.

I hope it is clear now, and no need to be terse.

Cheers,
can