Divers in Acapulco dive from a cliff that is 61 m high. If the rocks below extend outward for 23m what is the minimum horizontal velocity a diver must have to clear the rocks??
do you break into compents?? like
23cos(90)=0
23sin(90)=23
plz help my teacher is horrible!!
2007-11-27 09:38:07 · 3 answers · asked by softballviolinist14 3 in Science & Mathematics ➔ Physics
ok, the way to solve these is to work out how long you will fall for and then use that time to work out the horizontal velocity.
remember this one... s = ut + 1/2 at^2
so here s = height of cliff, u = 0, a = g and t is what we want...
61 = 1/2 * 9.81 *t^2 ... . I get t = 3.53 seconds.
Now you have to dive forward to cover 23 meters in 3.53 seconds - so your horizontal velocity must be 23/3.53 = 6.52 m/s
hth
2007-11-27 10:11:22 · answer #1 · answered by noisejammer 3 · 1⤊ 1⤋
The time t it takes to fall from heigh h is the time it takes to clear the rocks. So we have
t=sqrt(2h/g)
The horizontal component is distance s divided by time. We have
v=s/t
v=s/sqrt(2h/g)
v=s sqrt(g/2h)
v=23 sqrt(9.81/ (2 x 61))
v=6.5m/s or actually
v> 6.5m/s
2007-11-27 10:01:55 · answer #2 · answered by Edward 7 · 1⤊ 0⤋
Thats a calculus problem not a trig problem. I believe it involves the equation of a conic section .(a hyperbola?perabola?) sines and cosines shouldnt come into play
2007-11-27 10:04:29 · answer #3 · answered by stvc1961 2 · 1⤊ 1⤋