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
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Answer:
The acceleration of the object is
Explanation:
Given:
Initial velocity of object = 200 feet/second
Final velocity of object = 50 feet/second
Time of travel = 5 seconds
To calculate acceleration of the object we will find the rate of change of velocity with respect to time.
So, acceleration is given by:
where represents final velocity,
represents initial velocity and
is time of travel.
Plugging in values to evaluate acceleration.
The acceleration of the object is (Answer). The negative sign shows the object is slowing down.