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A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimate the maximum height the ball reaches. (Use 10 m/s2 as the acceleration of gravity.) 101 Incorrect: Your answer is incorrect.

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hamzaahmeds

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

H = 45 m

raphealnwobi

The maximum height of the ball is 45 m.

Let u represent the initial velocity and θ represent the angle. Hence:

Vertical component is 30 m/s, hence:

u*sinθ = 30  (1)

Horizontal component is 15 m/s, hence:

u*cosθ = 15   (2)

Divide eqn 1 by 2:

tan(θ) = 2

θ = 63.4°

u*sin(63.4°) = 30

u = 33.5 m/s

The maximum height (h) is:

[tex]h=\frac{u^2sin^2\theta}{2g} \\\\h=\frac{33.5^2*sin^2(63.4)}{2*10} =45\ m[/tex]

The maximum height of the ball is 45 m.

Find out more at: https://brainly.com/question/16906506

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