A 10000 car comes to a bridge during a storm and finds the bridge washed out. The 700 driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.0 above the river, while the opposite side is a mere 1.60 above the river. The river itself is a raging torrent 60.0 wide.
5 answers
Feet, meters? Kg or pounds? g is 9.8 m/s^2 or 32 ft/s^2 ????
The mass does not matter here.
Use
drop = (1/2) g t^2 in whatever units
to solve for t
t = sqrt (2 drop/g)
that is time in air
distance = initial speed u * t
so
60 whatever units = u * sqrt (2 drop/g)
u needed = 60 sqrt(g/{2 drop})
Use
drop = (1/2) g t^2 in whatever units
to solve for t
t = sqrt (2 drop/g)
that is time in air
distance = initial speed u * t
so
60 whatever units = u * sqrt (2 drop/g)
u needed = 60 sqrt(g/{2 drop})
-- For this question, I know how to find the initial speed. If it asks, "What is the speed of the car just before it lands safely on the other side?" how do i calculate that because it's not the same as final velocity right?
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you got u, the horizontal speed
v = g t
which is the vertical speed at landing
s = sqrt (u^2 + v^2)
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you got u, the horizontal speed
v = g t
which is the vertical speed at landing
s = sqrt (u^2 + v^2)
Units are Newtons and Meters
Questions are, How fast must the car be traveling to land safely on the other side? and What is the speed of the car just before it lands?