1. An object is thrown vertically upward from a helicopter that is hovering 30.0m above the ground. The initial velocity of the object was 20.0m/s.
a) Calculate the velocity with which the object hits the ground.
b) Calculate the time it took for the object to reach the ground

For part a, why can't I use the formula vf^2=vo^2+2ad? Because then it would be
vf^2=(20m/s)^2+2(-9.8m/s^2)(30m), and the right side of the equation will be negative, and it cannot be square rooted anymore
Or should I first calculate the time

2. An object is thrown vertically upward. If this object takes 5.30s to go up and down, what height did it reach?

For this one, I simply don't know how to do it... so if anyone can provide me with some explanations, that will be great

For the first problem check your signs - the object ends up 30 m below its starting point, so the "d" term is -30, not +30. By using +30 you have just proved that the object never reaches a point 30m above the helicopter.

For the second, use the equation:


d_f = \frac 1 2 a t^2 + v_0 ^2 t + d_0


where d_0 and d_f are both 0. Solve for v_0 . Then use that value in


v_f ^2 = v_0 ^2 +2 a d


where v_f is 0 (since the velociity of the object is 0 at its maximum height), and solve for d.

For the first question
Step1
Use v=u-gt to find the time up to the maximum height, where v=0, which gives
2sec.
Step2
Use s=ut+1/2gt^2 taking g as negative
And using t=2 to find height. Which gives 20m
Step3
Add 20 and 30 to find the height to the ground
Step4
Use v^2=u^2+2gh to find the velocity which gives 31.6m/s
Part b
Since you are provided with the to and fro time.
Divide the time given to find the time it took to reach the maximum heigh
Then use s=ut+1/2gt^2, where u=0 which gives h=35.1m

I thinj that there is wrong in the question .I had never listen about helicopter that throw things upward , so if you mean that it release it :

you use : vf^2 = vo^2+2ad where a=10m/s^2

and d = 30m

The distance from A to B is 400Km/m.A bus leave A at 4:55pm,second bus leave B at 3:05.Both bus travel at 115Km/h.At what point will the bus meet? Bus

Sorry 400km not 400km/h

Have you considered to draw a quick sketch?

Remember that:

Speed = \frac{Distance}{Time}

Get two equations and solve simultaneously, being careful at the variables that you use.

Is the time correct? Because, if one bus leaves at 4.55 pm and the other one later, which must be at 3.05 am then they'll meet when the bus hasn't yet started from B... :confused:

for the first question you can't use that formula because the displacement is not 30m, since the object is thrown vertically upwards from the helicopter the displacement must be >30. What u have to understand for this question is that when the object hit the ground it's velocity is 0m/s. For the second question, visualize the entire thing as a parabola (with a maximum value), but it says that it took 5.3 secs, but this is the entire distance (we only need half the parabola) so u need to divide the time by two to get the time it took to travel from the start to half way. Which is 2.65s, now u have your givens : a=-9.8m/s^2, V2=0 m/s, and time= 2.65s. Now u can sub this in the equation d= V2(T) + (1/2A * T^2) to get your displacement which I found was 34.4m. Hope that helps and gl

#1
a)Given: g=-9.9 m/s^2, v(initial)= 20 m/s, v(final)= 0 m/s
d= vf^2-vi^2 all over 2(g)so that makes it like: 0^2-20^2 / 2(-9.8)
the answer is going to be d=20.41 m. Add your answer to the 30 m given. You total height would now be
50.41m. Now, since you have your distance already, use the formula vf= square root of vi^2 + 2gd. It
will be like vf= square root of 0^2 + 2(-9.8)(-50.41) and the answer to that would be vf= 31.45 m/s.

b) For the time, just use the formula t= square root of 2d / g. You will substitute the given so it will
be t= square root of 2(-50.41) / -9.8. The answer will be t= 3.41 s. You can also derive the solution
for time from average velocity=d/t. Your derived formula would be t=d/v. In the first question, your
distance was 50.41 right? It would fall from that height. Then, get the average velocity. Since it
would start from the maximum height, the initial velocity must be 0. The final would be 31.45. 31.45 /
2 would get 15.73. So put that in the formula. t=50.41/ 15.73. The answer would be 3.21 also.

2. Given: g=-9.8 m/s^2, t=5.30, and since it would fall down, the vf would be 0. So here, you would get
time up and down. You would divide 5.30 and the answer would be 2.65. Use the formula vi= vf + gt.
Substitute and you will have 0+(-9.8)(-2.65). The answer would be 25.97. That would be you initial
velocity. Use that for the formula in distance which is d= vf^2-vi^2 all over 2(-9.8). The answer would
be 34.41 m. If you would also use the formula d= vi(t) + 1/2 (g)(t)^2, substitute: d= 0(2.65) + 1/2
(-9.8)(2.65)^2, the answer would also be 34.41 m. Using any other formulas would also result to this
answer like d= average velocity times time.

Hope this helps!

Jecaroon:

Wecome to AMHD! I see you're new here, so let me offer some tips:

1. Note that the problem you're responding to was posted over 3 years ago. Please check the dates before bothering to answer.

2. In general on this board we do not provide complete solutions to homework problems, but rather we provide a bit of guidance and let the student try to finish it themselves. If the poster still has difficulty, he can always post back and ask for further guidance. Given that the OP hasn't asked for additional help on this in over three years, I suspect he's already turned his homework in.

We Can Solve 1st question by using the formula v^2=u^2+2as.where a=9.8m/s^2& for calculate the time we use t=(V-U)/a