Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

So now we have an equation and unkown value.
for the thrown rock

for the dropped rock

solving both equation with the quadratic formula:

we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
The right answer to this question is A. a crest that is toppling over. When a surfer rides an ocean wave on her surfboard, she is actually riding on a crest. The crest is the point on a wave with the maximum value or upward displacement within a cycle.
D = distance between the cars at the start of time = 680 km
v₁ = speed of one car
v₂ = speed of other car = v₁ - 10
t = time taken to meet = 4 h
distance traveled by one car in time "t" + distance traveled by other car in time "t" = D
v₁ t + v₂ t = D
(v₁ + v₂) t = D
inserting the values
(v₁ + v₁ - 10) (4) = 680
v₁ = 90 km/h
rate of slower car is given as
v₂ = v₁ - 10
v₂ = 90 - 10 = 80 km/h