Question #124079
A particle is projected vertically upward with a velocity of 15 m/s from the point O. Find
the:
(i) maximum height reached.
(ii) time taken for it to return to O.
(iii) time taken for it to be 25 m below O.
1
Expert's answer
2020-06-28T18:51:44-0400

According to Newton's second law


mdvdt=Fm{dv\over dt}=F

By the Law of conservation of energy

K+P=constK+P=const

Let yy denote the height of the particle above the ground.

Let v(t)v(t) denote the velocity of the particle at time tt

mv022+0=mv22+mgy{mv_0^2\over 2}+0={mv^2\over 2}+mgy

mdvdt=mgm{dv\over dt}=-mg


v(t)=v0gtv(t)=v_0-gt

(i) maximum height reached. 


v=0,y=ymaxv=0, y=y_{max}

mv022+0=0+mgymax{mv_0^2\over 2}+0=0+mgy_{max}

ymax=v022gy_{max}={v_0^2\over 2g}

ymax=(15m/s)22(9.8m/s2)11.480 my_{max}={(15m/s)^2 \over2(9.8m/s^2)}\approx11.480\ m

(ii) Let t2=t_2= time taken for it to return to 0.0.  

When the partice reaches maximum height


v(t1)=0=v0gt1v(t_1)=0=v_0-gt_1

t1=v0gt_1={v_0\over g}

t2=2t1=2v0gt_2=2t_1={2v_0\over g}

t2=2(15m/s)9.8m/s23.061 st_2={2(15m/s)\over 9.8m/s^2}\approx3.061\ s

(iii) time taken for it to be 25 m below O.


v(t)=dydtv(t)={dy\over dt}

y(0)=0,v(t)=v0gty(0)=0, v(t)=v_0-gt

y=v(t)dt=(v0gt)dt=y(0)+v0tgt22=y=\int v(t)dt=\int(v_0-gt)dt=y(0)+v_0t-{gt^2\over 2}==v0tgt22=v_0t-{gt^2\over 2}

Given y=25 my=-25\ m


25=15t39.8t322,t30-25=15t_3-{9.8t_3^2\over 2}, t_3\geq0

4.9t3215t325=04.9t_3^2-15t_3-25=0

t3=15±(15)24(4.9)(25)2(4.9)=t_3={15\pm\sqrt{(-15)^2-4(4.9)(25)}\over 2(4.9)}=

=15±7159.8={15\pm\sqrt{715}\over 9.8}

Since t30,t_3\geq0, we take


t3=15+7159.84.259 st_3={15+\sqrt{715}\over 9.8}\approx4.259\ s

The particle will be 25 m below approximately 4.2594.259 seconds after start.



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