Shadow Height Rate in Related Rates Problem

[physicsman2]

Homework Statement

A sandbag is dropped from a balloon at a height of 60m when the angle of elevation to the sun is 30 degrees. Find the rate at which the shadow is at a height of 35 meters

Homework Equations

the position of the sandbag is given by s(t)=60-4.9t^2

The Attempt at a Solution

i tried to find the relation between the shadow's path and the position of the sandbag, but my efforts were futile.

 

[Dick]

The point below the sandbag, the position of the sandbag and the position of the shadow make a right triangle. Use trig.

 

[physicsman2]

i know that but i need help trying to solve it
i don't know how to relate the shadow to something else

 

[Dick]

One leg of your triangle has length s(t). You want to figure out the length of the other leg. One of the angles in your triangle is 30 degrees.

 

[physicsman2]

i got the hypotenuse to be 70 and b to be 61 since a would be 35 since the problem asks when the height is 35
is this right and if so, how would i go from there I am still having trouble

 

[Dick]

What's the relation between the vertical leg (s(t)) and the horizontal leg? Which trig function would be good to use? One of them is the ratio between the two legs.

 

[physicsman2]

im sorry if the problem doesn't make sense
s(t)=60-4.9t^2 is the instantaneous position of the sandbag in the air, not a leg.
I think that's what makes this problem so confusing
so to find the sides of the triangle, i did sin(30)/35= the hypotenuse and tan(30)/35=the base of the triangle since the problem says when the sandbag is 35 feet in the air, giving us the 35 to use.

 

[Dick]

s(t) is the position of the sandbag at time t. It's also the length of the vertical leg at time t. If h(t) is the length of the horizontal leg at time t, then yes, s(t)/h(t)=tan(30 degrees). So s(t)=tan(30 degrees)*h(t). The rates are then related by d/dt(s(t))=d/dt(h(t))*tan(30 degrees). The vertical rate is equal to the horizontal rate times tan(30 degrees). You need to find the vertical rate when s(t)=35 and then solve for the horizontal rate. The other lengths aren't important.

 

[physicsman2]

okay so would 60-4.9t^2=xtan(30) where x=horizontal length so if you differentiate that you would get -9.8t=(dx/dt)tan(30)+xsec^2(30)
is this right and if so where would i go from here
I am sorry for being a pain but i really don't understand this problem

 

[Dick]

Don't worry about being a pain, but no, you don't understand the problem. But you are getting closer. You have to figure out what the rate -9.8*t is at the point where 60-4.9*t^2=35. That means you have to solve the latter equation for t and plug it into the first equation to get the vertical rate. Once you have that, (vertical rate)=(horizontal rate)*tan(30). You don't have to differentiate the trig function, it's a constant!