Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?
The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m
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If the mass of one object is doubled, the force between these objects will also double.
Force refers to an influence on a body which can change its state of rest or motion.
Force F = G(mM/d^2) (where m is the mass of first object, M is the mass of the second object and G is the gravitational pull and d is the distance between the two objects)
The force between two objects (m and M) (according to the universal law of gravitation) is proportional to their mass and reciprocally proportional to the square of their separation (R) between them.
So F' = G(2mM/d^2),
which means F'=2F
Therefore, the when the mass is doubled, the force also doubles.
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Answer:
Final velocity of the first person is 3.43m/s and that of the second person is 0.0242m/s
Explanation:
Let the momentum of the first person, the ball second person be Ma, Mb and Mc.
From the principle of the conservation of momentum, sum of the momentum before collision is equal to the sum of the momentum after collision.
Ma1 + Mb1 = Ma2 + Mb2.
The ball and the first person are both moving together with a common velocity 3.45m/s.
Let the velocity of the first person be v1
Therefore
67.5×3.45+ 0.041×3.45= 67.5v1 + 0.041×34
233.02 = 1.39+ 67.5v1
67.5v1 = 233.02 - 1.39 = 231.61
v1 = 231.61 / 67.5
v1 = 3.43m/s
The second person and the ball move together with a common velocity after catching the ball.
For the second person and the ball let their final common velocity be v
Mb2 + Mc2 = Mb3 + Mc3
0.041 × 34 + 57.5 ×0 = (57.5 + 0.041)×v
57.541v = 1.39
v = 1.39 /57.541
v = 0.0242m/s
