Answer:
Explanation:
Using kinematics equations:
Use due to condition of distance traveled.
Solving second equation for time, there are two solutions. t=0 and
Use the expression in the first equation to have
Using trigonometric identities, you have the answer of the distance.
By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.
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
