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3 years ago

A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start

Physics

1 answer:

n200080 [17]3 years ago

3 0

Answer:

the initial velocity of the ball is 104.67 m/s.

Explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

$T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s$

Therefore, the initial velocity of the ball is 104.67 m/s.

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A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial spee

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Answer:

θ = sin⁻¹ $\sqrt{2gd}$

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = $\sqrt{2gd}$

θ = sin⁻¹ $\sqrt{2gd}$

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