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laiz [17]
2 years ago
7

A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the ve

locity of the ball is 20.0 m/s. What is the magnitude of the
FINAL VELOCITY when it strikes the ground? (Note: Ignore air resistance.)
Physics
1 answer:
Charra [1.4K]2 years ago
5 0

Answer: 28.1

Explanation:

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A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
As rotational speed increases, thrust____?
never [62]
Increases exponentially is your correct answer
6 0
3 years ago
Directions: Select ALL the correct answers.
nordsb [41]
Definitely D. The brakes on a bike rub against the wheel. Not sure about the others.
5 0
3 years ago
A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and v
neonofarm [45]
Total displacement along the length of mountain is given as
L = 235 m
angle of mountain with horizontal = 35 degree
now we will have horizontal displacement as
x = L cos35
x = 235 cos35 = 192.5 m
similarly for vertical displacement we can say
y = L sin35
y = 235 sin35 = 134.8 m
6 0
2 years ago
Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro
goldenfox [79]

Answer:

tension in rope = 25.0 N

Explanation:

  • Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
  • Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

       ∑F = F - W = 0

       so

       F = W

       so tension in rope = F  = T  = 25 N

8 0
2 years ago
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