Noah modeled a layer of the sun by showing how particles of rice move up and down in boiling water. Which layer of the sun did N
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The ratio of the height attended by the object thrown upward to that thrown at some angle θ with the vertical is 1.
Note: It is assumed that the mass of the first object is m and the mass of the second object is 2m. Also, the first object is thrown upward and the second object is thrown at some angle θ with the vertical.
Projectile motion: Any object that is thrown in the air is called a projectile and the motion described by it under gravity is called the projectile motion.
If the air resistance is neglected, then the acceleration due to gravity g is the same for all objects irrespective of their masses. It is given that both objects remain in the air for the same time period. So first calculate the time period for the objects when they are in the air.
Time period of the first object: The first object is thrown in the air upwards so from the second kinematics equation,
h1=u1*t1-(1/2)*gt1^2
where h1 is the height, t1 is the time, and u1 is the initial velocity for the first object.
When the object is not in air h1=0, so
0=u1*t1-(1/2)*gt1^2
After solving the above quadratic equation, the values of t1 obtained are t1=0 which represents the initial time, and t1=2u1/g which represents the time period. So the time period of the first object is,
t1=2u1/g
Time period of the second object: The second object is thrown at some angle θ with the vertical as shown in the diagram. From the diagram, the initial velocity along the vertical direction is,
u2=uocos(θ)
where uo is the initial velocity and u2 is the initial velocity along the vertical direction.
From the second kinematic equation,
h2=u2*t2-(1/2)*gt2^2
where h2 is the height, t2 is the time, and u2 is the initial velocity for the second object along the vertical direction.
When the object is not in air h2=0, so using u2=uocos(θ),
0=uocos(θ)*t2-(1/2)*gt2^2
After solving the above quadratic equation, the values of t2 obtained are t2=0 which represents initial time, and t2=2uocos(θ)/g which represents the time period. So the time period of the second object is,
t2=2uocos(θ)/g
Given that the time period is the same for both cases,
t1=t2
2u1/g=2uocos(θ)/g
u1=uo cos(θ)
Calculation of the ratio of the height of the object:
The maximum height is attained when the time of the object in the air is half of the total time period. At maximum height, velocity is zero.
From the third equation of motion.
v^2=u^2-2gh
h=u^2/2g
where h is the height and u is the initial velocity of an object.
Using it and u1=uo cos(θ) and u2=uo cos(θ), the ratio of h1 and h2 is,
h1/h2= u1^2/2g÷u2^2/2g
h1/h2=(uo cos(θ))^2/(uo cos(θ))^2
h1/h2=1
The ratio of their height will be 1.
Learn more about projectile motion.
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