Physics, 07.03.2020 05:15 londonval

A rock is tossed straight up from the ground with a speed of 19 m/s . When it returns, it falls into a hole 10 m deep.

a.) What is the rock's velocity as it hits the bottom of the hole?

b.) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

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Answer from: nonjabulomabaso7423

a) 23.6 m/s b) 4.35 s

Explanation:

Once tossed straight up, the rock is only accelerated by gravity, which is always directed downward.This acceleration, as is opposite to the movement of the rock, is slowing down it, until the instant that when reached to the heighest point, the rock comes momentarily to an stop.If we apply the definition of acceleration, to that moment, we can write the following equation:

$v_{f} = v_{0} -g*t_{h} = 0$

So, we can solve for the time needed to get to the heighest point, as follows:

$t_{h} = \frac{v_{0} }{g} = \frac{19 m/s}{9.8 m/s} = 1.94 sec. (1)$

Now, we can find the maximum height, applying the following kinematic equation:

$v_{f}^{2} - v_{0}^{2} = 2*g*h (2)$

If we choose the downward direction as the positive one, we can solve for h, as follows:

$h = \frac{v_{0} ^{2} }{2*g} = \frac{(19 m/s)^{2} }{2*9.8m/s2} = 18.4 m$

Now, from (1) taking the initial velocity as 0, we can find the velocity of the rock when returned to the level from where was tossed up, as follows:

$v_{f} = \sqrt{2*g*h} =\sqrt{2*9.8 m/s2*18.4m} = 19 m/s$

Now, we can apply (1) again, to get vf, the velocity of the rock at it hits the bottom of the hole, replacing v₀ = 19 m/s, and h = 10 m, as follows:

$v_{f} = \sqrt{v_{0}^{2} + (2*g*h)} =\sqrt{(19 m/s)^{2} + (2*9.8 m/s2*10m)} = 23.6 m/s$

The time during which the rock is in the air, can be divided in three parts:Time needed to reach to the maximum height = t₁ = 1.94 sec. (from (1)).Time needed to fall from this height to the level from where it was tossed up.Time needed to fall from this level to the bottom of the hole.In order to get the second time, we can use the equation for vertical displacement, replacing h = 18.4 m and v₀ = 0, as follows:

$h = \frac{1}{2} * g* t_{2}^{2} = 18.4 m$

Solving for t₂, we get:

$t_{2} = \sqrt{\frac{2*h}{g}} = \sqrt{\frac{2*18.4m}{9.8m/s2} } = 1.94s$

Not surprisingly, the time needed to reach to the maximum heigth is the same used to return to the level from which it started.In order to get the third time, we can apply the definition of acceleration, replacing vf = 23.6 m/s and v₀= 19 m/s, solving for t:

$t_{3} =\frac{v_{f} - v_{0}}{g} = \frac{23.6 m/s-19 m/s}{9.8 m/s2} = 0.5 s$

The total time that the rock is in the air is just the sum of the three times: t = t₁ + t₂ + t₃ = 1.94 s + 1.94 s + 0.47 s = 4.35 s
A rock is tossed straight up from the ground with a speed of 19 m/s . When it returns, it falls into

A rock is tossed straight up from the ground with a speed of 19 m/s . When it returns, it falls into

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