He leaves the slide at horizontal speed V and keeps that horizontal speed V for 5 meters and .5 seconds.
From that we can get speed V, because distance = speed times time.
Now if you know about conservation of energy and this slide is very slippery indeed we can say is kineic energy (1/2)mV^2 = m g h
that gives us the height of the slide itslf. HOWEVER we must add the distance fallen to the water. Remember the speed off the slide was horizonal, so there is no initial speed down.
Well how far does one fall in .5 seconds?
d = (1/2)9.8 t^2
so add those two heights
A water slide is constructed so that swimmers, starting from rest at the top of the slide leave the end of the slide traveling horizontally. There is still a height between the surface of the water and the end of the slide. One person hits the water 5.00m far from the end of the end of the slide in a time of 0.500s after leaving the slide. Ignoring friction and air resistance, find the height H of the water slide (from the top of the slide to the water surface).
Can anyone please give me some hints to do it?THANKS A LOT!
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