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Problem 44 Medium Difficulty

A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0 s later it is rising at a speed of 15 $\mathrm{m} / \mathrm{s}$ . Assuming air resistance has no effect on the rock, calculate its speed $(\mathrm{a})$ at launch and (b) 5.0 $\mathrm{s}$ after launch.

Answer

(a) $v_{0}=34.6 \mathrm{m} \cdot \mathrm{s}^{-1}$
(b) $v=-14.4 \mathrm{m} \cdot \mathrm{s}^{-1}$

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Physics 101 Mechanics

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Watch More Solved Questions in Chapter 2

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Video Transcript

Okay, So in this problem, we have a blast of dynamite sending a rock flying up from the sky. Um, part a is asking us to find its initial velocity, Given a few pieces of information that we know, those information are that it is travelling at 15 meters per second, two seconds after it begins rising upwards. And because it is a projectile motion problem meeting, there's an object in free fall in the air. The acceleration is 9.8 meters per second towards the grant meter per second squared towards the ground, which is in the negative direction. So just like that, I have three of my numbers that I know. The problem is asking me to solve for the initial velocity. So the only variable I don't know and don't want to know is X. So I'm gonna use the equation that does not have an accident which for this one is just our first equation. The V equals V not plus a t. When I rearrange this, we get that V not is equal to your final velocity minus acceleration versus time. And then when I plug in my numbers, we have 15 minus um 15 minus the acceleration of negative 9.8 times two. And the double negative there, when we put this into a calculator gives us our answer of 34.6 meters per second, which is how fast it was going initially when it left the ground. Now be asks us how fast we should expect it to be traveling five seconds after this explosion. So a few pieces there that we know first up the answer we got for a we're gonna carry over here because that answer of 34.6 is still the initial velocity of this chunk of rock as it's flying upwards, This is still a projectile motion. Problems are acceleration is negative, 9.8 meters per second squared. And lastly, we know the time frame is five seconds. So just like before, we're looking for the final velocity this time, but we don't know the X. We don't want to know the axe. So we're using the same equation we did in a When I plug in my numbers, we get 34.6 plus negative 9.8 multiplied by five and we get a final velocity of negative 14.4 meters per second. And if we were to translate this into English, that negative sign is just telling us that five seconds after the blast, this rock is moving 14.4 meters per second towards the

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