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2 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]2 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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If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll

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Explanation:

a) Given S(t) = 76 + 128t − 16t²

s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

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2 years ago

The velocity of an object is equal to the distance divided by time. The equation is velocity = distance/time. If you wanted to c

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Explanation:

The velocity $V$ equation is distance $d$ divided by time $t$ :

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Answer:

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We need to calculate the orbital speed of this second satellite

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From this relation,

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Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

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