Ok so use trigonometry to work out the vertical component of velocity.
sin(25) =opp/hyp
rearrange to:
30*sin(25) which equals 12.67ms^-1
now use SUVAT to get the time of flight from the vertical component,
V=U+at
Where V is velocity, U is the initial velocity, a is acceleration due to gravity or g. and t is the time.
rearranges to t= (V+u)/a
plug in some numbers and do some maths and we get 2.583s
this is the total air time of the golf ball.
now we can use Pythagoras to get the horizontal component of velocity.
30^2-12.67^2= 739.29
sqrt739.29 = 27.19ms^-1
and finally speed = distance/time
so--- 27.19ms^-1*2.583s= 70.24m
The ball makes it to the green, and the air time is 2.58s
Rain fall which is called run off
Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
