A plane drops a bomb from an altitude of 10,000m.
a. Determine the time the bomb is in the air.
b. Determine the velocity the bomb would have right before it hits the ground.
c. Before the bomb is launched the plane has a horizontal velocity of 92m/s. Determine the distance the bomb will travel in the X-axis.
d. Solve for the Resultant velocity that the bomb would impact the ground with.
Thanks for the help!
a. To determine the time the bomb is in the air, we can use the kinematic equation for vertical motion:
h = (1/2) * g * t^2
Where:
h is the vertical distance the bomb has fallen (10,000m),
g is the acceleration due to gravity (-9.8m/s^2),
t is the time in seconds.
Rearranging the equation to solve for time (t):
t = sqrt(2h/g)
Substituting in the given values:
t = sqrt(2 * 10,000 / 9.8) = sqrt(2040.82) = 45.17 seconds
Therefore, the bomb is in the air for approximately 45.17 seconds.
b. To determine the velocity the bomb would have right before it hits the ground, we can use the kinematic equation for vertical motion:
v = g * t
Where:
v is the velocity in m/s,
g is the acceleration due to gravity (-9.8m/s^2),
t is the time in seconds (which we have already determined as 45.17s).
Substituting the values:
v = -9.8 * 45.17 = -444.09 m/s (Note: the negative sign indicates the velocity is directed downwards)
Therefore, the bomb would have a velocity of approximately 444.09 m/s right before it hits the ground.
c. To determine the distance the bomb will travel in the X-axis, we need to consider the horizontal velocity of the plane. We can use the equation:
d = v * t
Where:
d is the distance in meters,
v is the horizontal velocity (92 m/s),
t is the time in seconds (45.17s).
Substituting the values:
d = 92 * 45.17 = 4147.64 meters
Therefore, the bomb will travel approximately 4147.64 meters in the X-axis.
d. To solve for the resultant velocity, we can use the Pythagorean theorem:
Resultant velocity = sqrt((velocity in x-axis)^2 + (velocity in y-axis)^2)
The velocity in the x-axis is 92 m/s (as given), and the velocity in the y-axis can be calculated using the vertical velocity obtained in part b (which is -444.09 m/s).
Resultant velocity = sqrt(92^2 + (-444.09)^2) = sqrt(8464 + 197100.2081) = sqrt(205564.2081) = 453.59 m/s
Therefore, the resultant velocity that the bomb would impact the ground with is approximately 453.59 m/s.