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docker41 [41]
3 years ago
12

two objects are thrown from the top of a tall building and experience no appreciable air resistance. one is thrown up and the ot

her is thrown down, both with the same initial speed. what are their speeds when they hit the street?
Physics
2 answers:
NNADVOKAT [17]3 years ago
5 0

Answer:

Final speeds are equal to each other.

Explanation:

Both objects experiment a change in their velocities because of gravity. Let model each object by following equations:

Object A - Throw Up

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Object B - Throw Down

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Where h is the height difference between the top of the building and the street. Let consider upward speed positive.

Final speeds are equal to each other.

Iteru [2.4K]3 years ago
3 0

Answer:

The two objects are traveling at the same speed.

Explanation:

Neglecting air resistance, an object that is thrown up from the top of a tall building has the same speed as the second object thrown down from the top of the same tall building since the initial speed is the same.

The object thrown up is not traveling faster neither is the object thrown down traveling faster.  

Therefore, the two objects will have the same speed when they hit the ground but their time of landing might be different.

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4 years ago
How does newton's first law of motion relate to galileo's concept of inertia?
Nikolay [14]
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3 years ago
At left A red ball in a box with arrows pointing away from the ball in all directions. In the middle, a blue ball in a box with
MAVERICK [17]

Answer:

first one is b 2nd one is a 3rd is c and the 4th one is c also

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6 0
2 years ago
Read 2 more answers
Two protons are released from rest when they are 0.750 {\rm nm} apart.
Alex787 [66]

Explanation:

Given:

m = 1.673 × 10^-27 kg

Q = q = 1.602 × 10^-19 C

r = 0.75 nm

= 0.75 × 10^-9 m

A.

Energy, U = (kQq)/r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9

v = 1.356 × 10^4 m/s

B.

F = (kQq)/r^2

F = m × a

1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2

a = 2.45 × 10^17 m/s^2.

4 0
3 years ago
<
goblinko [34]

Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

6 0
3 years ago
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