Answer:
I got you.. i'm in middle school and had that same question.
Explanation:
Refer to the diagram shown below.
The vertical distance traveled is
s = 25 m
The initial vertical launch velocity is zero.
Therefore
s = (1/2)*g*t²
where g = 9.8 m/s²
t = the time of flight, s
That is,
0.5*9.8*t² = 25
t² = 25/4.9 = 5.102
t = 2.26 s
Answer: 2.26 s
Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So

As we find acceleration.Now we need to find time
So

Now for distance
So

Answer:
A) 17.7 m/s
B) 15.98 m
C) Zero
E) 9.8 m/s²
Explanation:
given information
distance, h = - 34 m
time, t = 5 s
A) What is the initial speed of the egg?
h - h₀ = v₀t -
t², h₀ = 0
- 34 = v₀ 5 - \frac{1}{2} 9.8 5²
- 34 = 5 v₀ - 122.5
v₀ = 122.5 - 34/5
= 17.7 m/s
B) How high does it rise above its starting point?
v² = v₀² - 2gh
v = 0 (highest point)
2gh = v₀²
h = v₀²/2g
= 17.7²/2 (9.8)
= 15.98 m
C) What is the magnitude of its velocity at the highest point?
v = 0 (at highest point)
E) What are the magnitude and direction of its acceleration at the highest point?
g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.
Answer: 1300m
Explain: from km to m times 1000
Answer:
The correct option is A = adiant negy.