Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
Answer:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards
</em>
Explanation:
<u>Vertical Throw
</u>
When an object is thrown upwards, it describes a special type of motion ruled only by gravity.
When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.
The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.
With all these concepts in mind, we state that:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards </em>
The other options are not correct because:
A. The acceleration is never upwards
B. The acceleration is never 0
C. Both times are equal
Using trigonometric ratios we can get the distance;
For the first car; The distance from the point on the highway below the plane
tan = opp/adj
tan(36°) = 5150/x
0.727 = 5150/x
0.727x = 5150
x = 7088.37
For the second car we also use tangent; the distance from the point on the highway below the plane will be;
tan(56°) = 5150/y
1.483 = 5150/y
1.483y = 5150
y= 3473.72
The we can add the two distances to get how far apart the cars are;
7088.37 + 3473.72 = 10562.09 feet.
= 10562.09 ft
