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a particle is projected from ground in vertical direction at t = 0. At t = 0.8s, it reaches h = 14 m. It will again come to same height at t =

a particle is projected from ground in vertical direction at t = 0. At t = 0.8s, it reaches h = 14 m. It will again come to same height at t =

Grade:11

2 Answers

rahul chouhan
13 Points
2 years ago
FROM GROUND TO HEIGHT(T = 14m)
S=ut+1/2(-g)t^2
14=u0.8-1/2*10*(0.8)^2
[u=21.5m/sec]
speed at height = 14m
v=u+at
v=21.5-10*0.8
[v=13.5m/sec]
again,
we know that from B(H=14m @ t=0.8) to C(H=14m @ t=?) displacement is 0. 
i.e.
s=0 in,
            0=ut+1/2(-g)t^2
            [t=2.7 sec]
But this time is from B(h=14 @ t=0.8 sec) to C(h=14m @t=2.7 sec)
Question me he asked it will come to c from origin:
t=2.7+0.8
Ans.[T=3.5sec] 
Mustafizur Rahman
26 Points
one year ago
Here, s1= 14 m, t1= 0.8 sec, a1= -9.8 ms-2
Now,
s1      = u1t1+ (1/2)at12
14   =0.8u1  -  (1/2) x 9.8 x 0.82
0.8u1= 14 + 4.9 x 0.64
0.8u1= 14 + 3.136 = 17.136
      u1= 21.42 ms-1
Again,
v12  = u12  +  2as1
v12  =  458.8164 -  19.6 x 14 =  184.4164
v1  =  13.58  ms-1  or  -13.58  ms-1
Now,
v2           = -13.58
u1 +  at2  = -13.58
    -9.8 t2= -13.58 - 21.42 = - 35.00
           t2 = 3.57 sec

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