Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

as we know that




now from above equation we have



so the speed of combined system is 2 m/s
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Answer:
Speed of the speeder will be 28 m/sec
Explanation:
In first case police car is traveling with a speed of 90 km/hr
We can change 90 km/hr in m/sec
So 
Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m
After that car is accelerating with
for 7 sec
So distance traveled by car in these 7 sec

So total distance traveled by police car = 224 m
This distance is also same for speeder
Now let speeder is moving with constant velocity v
so 
v = 28 m/sec
Answer: 10.3m/s
Explanation:
In theory and for a constant velocity the physics expression states that:
Eq(1): distance = velocity times time <=> d = v*t for v=constant.
If we solve Eq (1) for the velocity (v) we obtain:
Eq(2): velocity = distance divided by time <=> v = d/t
Substituting the known values for t=15s and d=155m we get:
v = 155 / 15 <=> v = 10.3