The most pressing problem in the Chesapeake Bay is
2 answers:
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Answer:
13 m/s east
Explanation:
We can solve the problem by using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:
where
m = 0.1 kg is the mass of each puck
u1 = +13 m/s is the initial velocity of puck 1
u2 = -18 m/s is the initial velocity of puck 2 (here I assume the west direction to be the negative direction, so I put a negative sign)
v1 = -18 m/s is the final velocity of puck 1
v2 = ? is the final velocity of puck 2
Simplifying m from the formula and substituting the data, we can find the final velocity of puck 2, v2:
And the positive sign means that puck 2 is moving east.
Answer:
The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.
Explanation:
Given;
vertical height of the cliff, h = 22 m
velocity of the ball, u = 18 m/s at an angle 39⁰
vertical component of the velocity,
The time for the ball to get to the bottom of the cliff is calculated as;
h = ut + ¹/₂gt²
h = (u sinθ)t + ¹/₂ x 9.8 x t²
22 = (18 sin 39)t + 4.9t²
22 = 11.328t + 4.9t²
4.9t² + 11.328t - 22 = 0
Solve the above equation with formula method;
a = 4.9, b = 11.328, c = -22
The time for the player to get to the bottom of the cliff;
Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;
The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.