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Serhud [2]
3 years ago
10

someone throws a rubber ball vertically upward from the roof of a building 8.2m in height. the ball rises, then falls.it just mi

sses the edge of the roof, and strikes the ground.if the ball is in the air for 6.2s, what was its initial velocity?(disregard air resistance. a=-g=-10m/s²
Physics
1 answer:
laila [671]3 years ago
5 0

Answer:

H = V t - 1/2 g t^2         V and g are in different directions

-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2

V = 184 / 6.2 = 29.7 m/s

Check: find time for ball to return to zero height

0 = 29.7 T - 5 T^2

T = 29.7 / 5 = 5.94 sec

The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec

S = 29.7 * .26 + 5 * .26^2 = 8.1

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In the periodic, the elements are in order based on
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Question 8 of 10
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3 years ago
Sorry I’ve just been having trouble with this question
tiny-mole [99]

Answer:

68.8 N 13.8°N of W

Explanation:

F₁ is 50 N 30°N of W.  The terminal angle is 150°.

F₂ is 25 N 20°S of W.  The terminal angle is -160°.

Graphically, you can add the vectors using head-to-tail method.  Move F₂ so that the tail of the vector is at the head of F₁.  The resultant vector will be from the tail of F₁ to the head of F₂.

Algebraically, find the x and y components of each vector.

F₁ₓ = 50 N cos(150°) = -43.3 N

F₁ᵧ = 50 N sin(150°) = 25 N

F₂ₓ = 25 N cos(-160°) = -23.5 N

F₂ᵧ = 25 N sin(-160°) = -8.6 N

The x and y components of the resultant vector are the sums:

Fₓ = -43.3 N + -23.5 N = -66.8 N

Fᵧ = 25 N + -8.6 N = 16.4 N

The magnitude of the resultant force is:

F = √(Fₓ² + Fᵧ²)

F = √((-66.8 N)² + (16.4 N)²)

F = 68.8 N

The direction of the resultant force is:

θ = tan⁻¹(Fᵧ / Fₓ)

θ = tan⁻¹(16.4 N / -66.8 N)

θ = 166.2°

θ = 13.8°N of W

6 0
3 years ago
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