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Marta_Voda [28]
3 years ago
12

A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s

trikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed ? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).
Physics
1 answer:
ICE Princess25 [194]3 years ago
5 0

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

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Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with
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y₀ = 10.625 m

Explanation:

For this exercise we will use the kinematic relations, where the upward direction is positive.

         y = y₀ + v₀ t - ½ g t²

in the exercise they indicate the initial velocity v₀ = 8 m / s.

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V=xf-xi/t solve for t
irina [24]

Answer:

t=\frac{x_f-x_i}{v}

Explanation:

Starting from the equation:

v=\frac{x_f-x_i}{t}

First of all, let's multiply by t on both sides:

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And then, let's divide by v on both sides:

\frac{vt}{v}=\frac{x_f-x_i}{v}\\t=\frac{x_f-x_i}{v}

So, finally

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