Finding the Y-Coordinate of a Rhino at Time t=11.0 s

[badman]

A rhinoceros is at the origin of coordinates at time t_1=0. For the time interval from t_1 to t_2 = 11.0 s, the rhino's average velocity has x-component −3.40 m/s and y-component 5.00 m/s.

it asks for At time t_{2}= 11.0 s what is the y-coordinate of the rhino?


i already found the x coordinate which is -37.4 now i tried pluging in the initial velocity of the y coordinate 5 m/s. into this equatioon

distance= initial postion( which is zero)+ initial velocity times the time(which is equals to 55m)+ 1/2 times 9.81*t^2

if the acceleration is -, then the answer should be -538m
if its not then its postive then it should be 648m. but it says I am wrong on both accounts.

 

[AKG]

You don't know how the object is accelerating, why do you assume it's free-falling? Also, that equation you used only works for constant acceleration, and of course, you don't know how it's accelerating. Moreover, "initial velocity times time" is not 55m. 55m is "average velocity times time." You're given average velocity, not initial velocity. What's good about this is that you don't even need to know it's acceleration. Do you know what average velocity is? It's total distance over total time. Multiply this by the total time, and you get total distance, which is what you want.

 

[badman]

ahhhh i was thinking to hard on that problem.

 

[badman]

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \alpha of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.

these are the answers I've already found.

Find the angle of elevation \alpha.
53.1 ^\circCorrect

Part B
Find the speed of the water at the highest point in its trajectory.
15.0 m/sCorrect

Part C
Find the magnitude of the acceleration of the water at the highest point in its trajectory.
9.80 {\rm m/s^2}Correct

Part D
How high above the ground does the water strike the building?
15.9 mCorrect

Part E
How fast is it moving just before it hits the building?


im stuck on this one^

 

[whozum]

Identify your initial velocity vector, and break it up into components. The x component has no forces acting on it thus remains constant, but your y component is being accelerated by gravity. Find the effect of gravity on the y component to find your final v_y, then recombine the two components to find the final velocity.

 

[badman]

can you hint me on trying to find the initial celocitry vector?

 

[whozum]

You are given the initial launch angle and initial launch velocity. Draw a triangle and use some trig properties to find the components.

An easier way to do this would be to start when the water is at its highest point, above you said its horizontal velocity is 15m/s, and you know its vertical velocity is 0.

Velocity vector:

\vec{v}(t) = <15, gt>

Where t is the time it takes for the water to reach the building from its highest point.

 

[badman]

thanks man, i hope I am not annoying you.

 

[whozum]

Did you get the final answer?

 

[badman]

yes i did its 17.7 m/s thanks alot. i just got the two velocity compenents. and raised each to the second power, added them, then sqare'em.