Answer:
0.66c
Explanation:
Use length contraction equation:
L = L₀ √(1 − (v²/c²))
where L is the contracted length,
L₀ is the length at 0 velocity,
v is the velocity,
and c is the speed of light.
900 = 1200 √(1 − (v²/c²))
3/4 = √(1 − (v²/c²))
9/16 = 1 − (v²/c²)
v²/c² = 7/16
v = ¼√7 c
v ≈ 0.66 c
Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
PART a)
here when stone is dropped there is only gravitational force on it
so its acceleration is only due to gravity
so we will have

Part b)
Now from kinematics equation we will have

now we have
y = 25 m
so from above equation


Part c)
If we throw the rock horizontally by speed 20 m/s
then in this case there is no change in the vertical velocity
so it will take same time to reach the water surface as it took initially
So t = 2.26 s
Part D)
Initial speed = 20 m/s
angle of projection = 65 degree
now we have




PART E)
when stone will reach to maximum height then we know that its final speed in y direction becomes zero
so here we can use kinematics in Y direction



so it will take 1.85 s to reach the top
Answer:
<em>Time period of pendulum is 2.02 s.</em>
Explanation:
A <em>simple pendulum</em> is a device which consists of mass m hanging from the string of length L attached to the some point.When displaced and released its swings back and forth with periodic motion.
The time period of pendulum is defined as time taken by the pendulum to complete one full oscillation . it is denoted by T.
By <em>Huygens law of period of pendulum</em>,
T = 2π
eqn 1
where L is the length of pendulum,
g is acceleration due to gravity
<em>Period of pendulum is independent of the mass of pendulum,</em>
<em />
Substituting values in eqn 1
T = 2π 
T = 2.02 s
<em>Time period of pendulum is 2.02 s.</em>
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