Physics, 21.12.2019 09:31 Fangflora3
Abody falls freely from rest. find (a)its accleration (b) the distance it falls in 3s (c) its speed after falling 70m (d) the time requied to reach a speed of 25 m/sec (e) the time taken to fall 300m.
Answers: 1
![y = y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y_{0} =initial position =0[m]\\v_{0}=initial velocity = 0[m/s]\\y= distance traveled [m]\\t=time = 3[s]\\y = 0 +0 *(3)+\frac{1}{2} *(9.81)*(3)^{2} \\\\y=44.14[m]](/tpl/images/0428/9019/d3040.png)
![v_{f} ^{2} = v_{0} ^{2} +2*g*y\\where:\\v_{f}= final velocity [m/s]\\v_{0} = initial velocity [m/s]\\g = gravity [m/s]\\y=distance traveled[m]](/tpl/images/0428/9019/f47f5.png)
![v_{f} ^{2} =0+2*9.81*(70)\\v_{f}=\sqrt{1373.4} \\v_{f}=37.05[m/s]](/tpl/images/0428/9019/d9202.png)
![v_{f} =v_{0}+(g*t )\\v_{0}=0[m/s]\\g=9.81[m/s^2]\\t = time[s]\\v_{f} =25[m/s]](/tpl/images/0428/9019/2ebc4.png)
![25=0+(9.81*t)\\t=2.55[s]](/tpl/images/0428/9019/ff355.png)
![y=y_{0} +v_{0} *t+\frac{1}{2} * g*t^{2} \\where:\\y = distance traveled=300[m]\\v_{0}=0[m/s]\\g=9.81[m/s^2]\\300=(0)*t+\frac{1}{2} *(9.81)*t^{2} \\t=\sqrt{\frac{2*300}{9.81} } \\t=7.82[sg]](/tpl/images/0428/9019/bdaa2.png)
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Abody falls freely from rest. find (a)its accleration (b) the distance it falls in 3s (c) its speed...
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