Answer:
The water should be released a distance of 1289.88 feet before the plane is on top of the bush fire.
Explanation:
Let's first find the time required for the water to fall down 390 ft onto the bush fire.
Initial Vertical Speed of water = 0
Distance to be covered = 390 ft
Acceleration due to gravity = 32.2 ft/s^2
t = 4.92 seconds
Thus the water should be dropped 4.92 seconds before the plane is over the bush fire. Now we can also find the distance d at which the pilot should release the water:
d = Speed of plane * time
Speed of plane = 180 / (60 * 60) = 0.05 mile/second
d = 0.05 * 4.92 = <u>0.246 miles</u> OR <u>1289.88 feet</u>
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Thus, the water should be released a distance of 1289.88 feet before the plane is on top of the bush fire.
Answer:
I think the mean might be 6.7.
May you please make me brainliest?? that is if you want to tho.
Answer:
Multiply the acceleration by time to obtain the velocity change:
Example:
Velocity change = 6.95 * 4 = 27.8 m/s . Since the initial velocity was zero, the final velocity is equal to the change of speed. You can convert units to km/h by multiplying the result by 3.6: 27.8 * 3.6 ≈ 100 km/h
Answer:
I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.
Explanation:
In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:
R = V₀² Sin 2θ/g
where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.
The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:
V₀ = √(V₀ₓ² + V₀y²)
where,
V₀ₓ = Horizontal Velocity
V₀y = Vertical Velocity
Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.
<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>
Answer:Given:
Initial speed of fugitive, v0 = 0 m/s
Final speed, vf = 6.1 m/s
acceleration, a = 1.4 m/s^2
Speed of train, v = 5.0 m/s
Solution:
t = (vf-v0)/a
t = (6.1-0)/1.4
t =4.36 s
Distance traveled by train, x_T =v*t
x_T =5*4.36 = 21.8 m
Distance travelled by fugitive, x_f = v0*t+1/2at^2
x_f = 0*4.36+1/2*1.4*4.36^2
x_f =13.31 m
5*t = v(t-4.36)+x_f
5*t=6.1*(t-4.36)+13.31
solve for t, we get
t = 12.08 s
The fugitive takes 12.08 s to catch up to the empty box car.
Distance traveled to reach the box car is
X_T = v*t
X_T = 5*12.08 s
X_T = 60.4 m
Explanation:
