hf=hi+vi*time-4.9t^2
1.8=21.9+vi*t-4.9t^2
and
58=vi*t
solve for t in the second equation
t=58/vi
put that in the first equation, solve for vi
Now, speed just before impact
vertical speed=9.8*t
speedtotal=sqrt(vi^2+verticalspd^2)
During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 21.9 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 58.0 m wide.
A) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands on the other side?
2 answers
Bob seems to be using vi both for vertical and horizontal speed
I think in fact the original vertical speed is zero so
h = Hi + 0 -4.9 t^2
1.8 = 21.9 - 4.9 t^2
t^2 = 4.1
t = 2.03 seconds to fall
now the car has to go horizontal
58 meters in 2.03 seconds
so I will call the constant horizontal speed u
where
u = 58/2.03 = 28.6 m/s
then
speed = sqrt (28.6^2 + v^2)
where v = 0 + 9.81*2.03
v = 19.9 m/s
so
speed at landing = 34.9 m/s
by the way, that car is toast due to the 35 m/s vertical speed at landing
I think in fact the original vertical speed is zero so
h = Hi + 0 -4.9 t^2
1.8 = 21.9 - 4.9 t^2
t^2 = 4.1
t = 2.03 seconds to fall
now the car has to go horizontal
58 meters in 2.03 seconds
so I will call the constant horizontal speed u
where
u = 58/2.03 = 28.6 m/s
then
speed = sqrt (28.6^2 + v^2)
where v = 0 + 9.81*2.03
v = 19.9 m/s
so
speed at landing = 34.9 m/s
by the way, that car is toast due to the 35 m/s vertical speed at landing
2 answers