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
P = m*v
7.5 = m*15
m = 7.5/15 = 0.5 kg
Answer:
608kg
Explanation:
Formula : <u>Kinetic</u><u> </u><u>energy</u><u> </u>
½ ×mass x speed²
<u>47500</u>
½×12.5²
=608 Kg