A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o
f 3.24 km. The bombardier releases one bomb. (a) How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance.
1 answer:
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
You might be interested in
Answer:
Electrical force, F = 90 N
Explanation:
It is given that,
Charge on sphere 1,
Charge on sphere 2,
Distance between two spheres, d = 6 cm = 0.06 m
Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,
F = 90 N
So, the electrical force between them is 90 N. Hence, this is the required solution.