Two bullets are fired simultaneously parallel to a horizontal plane. The bullets have different masses and different initial vel
1 answer:
Answer:
Therefore, the answer is E. They strike the plane at the same time.
Explanation:
Here, it is seen that the time depends only on acceleration due to gravity (which is a constant) and vertical displacement, and not on velocity of the bullets or mass of the bullets.
Hence, the bullets that are fired simultaneously parallel to the horizontal plane will strike the plane at the same time.
using equation of motion for displacement
s= ut + ¹/₂gt²
here, g is the acceleration due to gravity along y- direction
U along y is 0
s = (0)t + ¹/₂gt²
s=¹/₂gt²
make t the subject of formula =
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Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.