Answer:
it would take 3.26 seg for the stone to fall to the water
Explanation:
If we ignore air friction then:
h=h₀ + v₀*t -1/2*g*t²
where
h= coordinates of the stone in the y axis ( height of the stone relative to the surface of the water )
h₀ = initial coordinates of the stone ( height of the cliff relative to the surface of the water = 52 m )
v₀ = initial <u>vertical </u>velocity = 0 ( since the ball is kicked horizontally , has only initial horizontal velocity , and has 0 vertical velocity )
t = time to reach a height h
g = gravity = 9.8 m/s²
since v₀ =0
h= h₀ - 1/2*g*t²
h₀ - h = 1/2*g*t²
t= √[2(h₀ - h)/g]
when the stone hits the ground h=0 ( height=0) , then replacing values
t=√[2(h₀ - h)/g]=√[2(52 m- 0 m )/(9.8m/s²)] = 3.26 seg
t= 3.26 seg
it would take 3.26 seg for the stone to fall to the water
Note that Speed = Distance/Time.
1st trip:
Distance = 30 mi
Time = 30 min = 0.5 h
Speed = 30/0.5 = 60 mi/h
2nd trip:
Distance = 50 mi
Time = 1 hour
Speed = 50 mi/h
3rd trip:
Distance = 30 mi
Time = 30 min = 0.5 h
Speed = 30/0.5 = 60 mi/h
Average speed = (60 + 50 + 60)/3 = 56.7 mph
The closestanswer is 55 mph
Answer: 55 mph
A) a mouse, to an order of magnitude = 0.1 m ( a tenth of a meter ) That would be a big mouse but the alternatives are 1 meter or one hundredth of a meter... so go with 1/10th
<span>b) Easy = 1 meter </span>
<span>c) two choices 10m or 100 m . Go with 100 m </span>
<span>d) Stretch it out , trunk tip to tail tip - call it 10 m </span>
<span>e) Your choice 100 m or 1000 m..... These are estimates. So long as you are within one order of magnitude you can't really be given wrong. So I'd say 100m</span>
