Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
When is at the end of the runway the velocity of the plane is given by the equation
where s=1800 m is the runway length. Thus
At half runway the velocity of the plane is
Therefore at midpoint of runway the percentage of takeoff velocity is
‰
Answer: 1) a = 9.61m/s² pointing to west.
2) (a) Δv = - 37.9km/s
(b) a = - 6.10⁷km/years
Explanation: Aceleration is the change in velocity over change in time.
1) For the plane:
a = 9.61m/s²
The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.
2) Total change of velocity:
km/s
The interval is in years, so transforming seconds in years:
v = 
km/years
Calculating acceleration:
Acceleration of an asteroid is a = -6.10⁷km/years .
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V
You should charge it to 1000 volts to store 1.0 J of energy.
Its 30 kg cause I got 30 kg dude.
