Basically my teacher hands us out a review packet for physics and says to do it if we want practice...Okay I do the packet and I have it done, now my problem is that the teacher didn't give out the answers to the problems so I have no idea whether or not I did it right...so Im hoping some one will check my work before I take my midterm tomorrow. Sorry for being so troublesome.(adsbygoogle = window.adsbygoogle || []).push({});

Anyways here are a few problems and my work involved. I don't know how to do those picture equations so bear with me

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1. The height of a helicopter above ground is given by h = 3t³, where h is in meters and t is in seconds. After 2 seconds, the helicopter releases a small mailbag, How long after its release does the mailbag reach the ground.

h = 3t³

h(2) = 24 m

h' = 9t²

v(2) = h' = (9)(2)² = 36 m/s

Si 0

Sf 24m

Vi -36 m/s

Vf 42 m/s

a 9.8 m/s²

t ?

I know that I have my signs placed oddly but thats how my brain works. Also I put all the information I have in the chart already from the work I get below except time.

Vf² = Vi² + 2a(Sf - Si)

Vf = sqrt(Vi² + 2a(Sf-Si))

Vf = sqrt((-36)² + (2)(9.8)(24))

Vf = 42 m/s

Vf = Vi + at

t = Vf-Vi/a

t = 42+36/9.8

t = 7.96 s

I think thats right but I wanna make sure.

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2. A 100 g bullet is fired from a rifle having a barrel 0.6 m long. Assuming the origin is placed where the bullet begins to move, the force (in Newtons) exerted on the bullet by the expanding gas is 15000 + 10000x - 25000x², where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

work:

F(x) = 15000 + 10000x - 25000x²

F(.6) = 15000 + 10000(.6) - 25000(.6)²

F(.6) = 12000 N

w = Fd

w = (12000N)(.6N) = 7200 J

Okay I think I did that problem horribly wrong because I mean its too easy and this is advanced physics.

One last problem

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A catcher "gives" with the ball when he catches a 0.15 kg baseball moving at 25 m/s. If he moves his glove a distance of 2 cm, what is the average force acating on his hand?

Work

Ke = (1/2)mv²

Ke = (1/2)(.15)(25)²

Ke = 46.87 J

Ke = (1/2)(.15)(0)² (Im assuming the ball stops)

Ke = 0 J

w = Kef - Kei

w = Fd

F = w/d

F = 46.87J/.02m = 2343.75 N

thanks for your time...I feel really bad now asking someone to check my work.

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# Mid-term tomorrow, need some checking of answers of review packet

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