Question

Please help. Will FAN AND MEDAL!

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the firework is 40 feet above the ground.

Answer 1

can you please help @Photon336

Answer 2

Where did the graph go that you just put up?

Answer 3

Okay I don't understand?

Answer 4

I'm guessing we could find a way to express theta the angle of elevation as a function of time.

but we know that from our physics equations that the change in the vertical distance is equal to

\[V_{o}Sin(\theta)t+\frac{ 1 }{ 2 }at^{2} = {y-y_{0}}\]

while the change in horizontal distance is equal to this.

\[V_{o}Cos(\theta)t = (x-x_0)\]

to find theta we know that if we know what the y distance traveled over x right and we take the inverse tangent we can find the angle theta.

\[\theta = tangent^{-1}\frac{ y }{ x }\]

\[\theta = tangent^{-1}(\frac{ y-y_{0}}{ x-x_{0} })\]

\[\theta = tangent^{-1}\frac{ V_{o}Sin(\theta)+\frac{ 1 }{ 2 }at^{2} }{ V_{o}*Cos(\theta)t }\]

\[\frac{ d~\theta }{ dt} = \frac{ d~\theta }{ dt } (tangent^{-1}\frac{ \cancel\V_{o}Sin(\theta)+\frac{ 1 }{ 2 }at^{2} }{ \cancel\V_{o}*Cos(\theta)t })\]

omg this is really nasty

\[\frac{ d~\theta }{ dt} = \frac{ d~\theta }{ dt } (tangent^{-1}\frac{ Sin(\theta)+\frac{ 1 }{ 2 }at^{2} }{ Cos(\theta)t })\]

@Kainui please help lol i'm lost.

Answer 5

@ganeshie8

Answer 6

I think we should assume that the fire cracker is launched vertically up

Answer 7

But then the firework won't go that high. Something seems to be wrong with the question here...

Answer 8

I think this is not a well thought out calculus problem where gravity doesn't exist