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Hey Key Set
throwing key question please help!?
Hey everyone. Hoping you can help me tonight. I am having some trouble with these and I was wondering if you could show me how to solve them. First, I estimated that the position function is -32t+55. I assume this is right? If so, is number 3 below simply setting -32t+55 to equal 25 and solving for t?
Standing on a church balcony 30 ft up from ground, you are throwing your key to a friend who is standing on the ground. You throw it angled upward with intilial velocity of 55 ft/sec. Acceleration from gravity is -32 ft/sec/sec
How many seconds does it take for the key to reach the ground, not your friend's hand?
How long has it been if the key is 40 feet in the air?
When is the velocity 25 ft/sec?
Any help is appreciated!
The best way to answer a question such as this is to break it into two questions. You need to know the total time it takes for the key to reach the ground, but before we tackle that, we need to know how long it takes before the key stops going up.
So what do you know for the up question?
Vo = 55fps
Vf = 0fps (the key stops at the top of its trajectory -- before it starts moving down)
a = -32ft/s^2
Look at your kinematic equations. As a refresher, they are:
d=Vo*t+0.5*a* t^2
Vf^2 = Vo^2 + 2*a*d
Vf = Vo + a*t
d = 0.5*(Vf + Vo)*t
Pick the equation that has the information you have and addresses the problem you need to solve. In this case, we'll use the 3rd equation:
Vf = Vo + a*t rewrite it as:
0=55+ (-32*t) subtracting 55 from both sides gives us:
-55=-32*t dividing both sides by -32 give us:
1.719 = t So it takes 1.72 seconds for the key to reach the highest point in its arc.
To do the second half of the problem, we need to know how high the key traveled during those 1.72 seconds. Look back at the kinematics and pick the best equation. The easiest one would be the last one
d = 0.5*(Vf + Vo)*t rewrite it as:
d = 0.5*(0+55)*1.719 solving for d we find that:
d = 47.273 ft
The key started at 30 ft above the ground. Now it's 30+47.3 feet above the ground (or 77.3 ft).
Now solve the downward problem. We know:
Vo = 0 (remember, we're at the top of the trajectory where the key stops for a fraction of a second)
d = 77.27 ft
a = 32 ft/s^2 (yes, I know it's supposed to be negative, but we'll call down positive since that's the direction the key is going. That makes the acceleration positive since it's acting in the same direction as motion)
Looking at the equations, that leaves us with d=Vo*t+0.5*a*t^2
Rewrite the equation like this:
77.27 = 0*t + 0.5(32)(t^2) That boils down to:
77.27/(0.5*32)=t^2
t^2 = 4.829
and so, t = square root(4.829) = 2.197 seconds
Ok, so it took 1.719 seconds for the key to reach the top of its arc and 2.197 seconds for it to go from the top of the arc to the ground.
Therefore, the key was in the air for a total of 3.916 seconds.
You will solve the other questions in a similar fashion. The only difference, is you can find two times for each answer (the key will be at 40ft on its way up and again on its way down -- same for the velocity). As a hint, solve for the velocity problem first because you know Vo, Vf, and acceleration, so you'll just need to solve for time using the third equation above.
I hope this gives you a better understanding on how to tackle this sort of problem and a good start at completing this assignment.
Set Driven Key Tutorial