A sled with rider having a combined mass of 120 kg travels over the perfectly smooth icy hill shown in the accompanying figure. How far does the sled land from the foot of the cliff (in m)?

sjeikdom0 2021-01-24 Answered

A sled with rider having a combined mass of 120 kg travels over the perfectly smooth icy hill shown in the accompanying figure. How far does the sled land from the foot of the cliff (in m)?
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au4gsf
Answered 2021-01-25 Author has 95 answers
The rider begins with
KE=12mv2=0.5(120kg)(22.5ms)2=30375 J
and
PE=mgh=1209.80=0
so,
Et=KE=30375 J
where
KE=kinetic energy
PE=potential energy
E=total energy=KE+PE
At the top of the hill some of the KE is turned into PE=mgh
PE becomes
PE=120kg9.8ms211.0m=12936 J
and since total energy is conserved KE becomes
KE=EtPE=30375 J12936 J=17439 J
KE=12mv2v=2KEm=217439120=17.05 ms
Now since the hill is 11.0 m high the period of time of thevertical fall is given by
h=v0t+12at2
h=11.0m
v0=0 (this is the initial vertical velocity)
a=9.8m/s
solve for t using the quadratic formula
t=va±v024(12a)(h)2(12a)
t=±2aha
choose the positive answer, as the negative answer for time is meaningless
t=2(9.8)(11)9.8=1.498s1.5 sec
so we know it takes 1.5 sec to hit the ground and we know the horizontal velocity is v=17.05m/s, distance travelled before impact:
d=vt=17.05ms 1.5s=25.5m
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