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coldgirl [10]
3 years ago
10

A basketball is thrown horizontally with an initial speed of

Physics
1 answer:
storchak [24]3 years ago
8 0

Answer:

1.35 m

Explanation:

Taking down to be positive, given:

Δx = Δy / tan 30.0º

v₀ₓ = 4.50 m/s

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = 10 m/s²

Find: Δy

First, find the time it takes to land in terms of Δy.

Δy = v₀ t + ½ at²

Δy = (0 m/s) t + ½ (10 m/s) t²

Δy = 5t²

Next, find Δx in terms of t.

Δx = v₀ t + ½ at²

Δx = (4.50 m/s) t + ½ (0 m/s) t²

Δx = 4.50t

Substitute:

Δy = 5 (Δx / 4.50)²

20.25 Δy = 5 (Δx)²

4.05 Δy = (Δx)²

4.05 Δy = (Δy / tan 30.0º)²

4.05 Δy = 3 (Δy)²

1.35 = Δy

The basketball was thrown from an initial height of 1.35 m.

Graph: desmos.com/calculator/ujuzdo9xpr

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Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

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Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

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Answer:

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Explanation:

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