Taking oringin at the top of building and upward direction positive.
For particled A `u=+10 s^(-1) a=-10 m s^(-1)`
Using `y=ut+(1)/(2)at^(2)`
`=10t-(1)/(2) xx10xxt^2) =10t-5^(2)` ...(i)
Equation (i) is the equation of parabola open down The parabola passes origin when `y=0`. hence from
`0=10t-5t^(-2) rArr t=0,0 s`
For velocity-time relation, `v=u+at`
`v=10-10t`. (ii)
The relaation is straight line relation with negative slopen and positive intercept.
At maximum hetght, velocity of the particle is zero.
`0=10-10 t rArr t=1 s`
Maximum height from (i), `y_(max)=10xx1-5xx1^(2)=5 s s^(-1)`
It will reach ground wher `y=-100 m`
Using `i` again, `-74=10t-5t^(2) rArr (t-5)(t+3)=0`
When given `t=5 s`, i.e., the particle will hit ground at `t=5 s`.
Velocity at the time of hitting
Using (ii), `v=10-10xx5=40 m s^(-1)`
For speed-time relation, the speed is always positive the mirror-image of velocity-time relation on positive side.
For drawing acceleration-time graph we have taken downward direction as negative and acceleration is constant throughout the motion of the praticle, i.e. `-10 m s^-(2)` Hence, acceleration time graph will be straight line parallel with time axis and below the time axis.
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d. For particle B:`u=-10 m s^(-1) a=-20 m s^(-2)`
Displacement-time relation,
`y=-10t-(1)/(2)xxt^(2)=-10t^(2)` ...(iii)`
It is a parabla open down and it passes through origin. The particel will reach ground when `y=-75 m`
From (iii) `=-10t-5t^(2)`
or `t^(2)+2t-15=0 rArr t=3 s`
The velocity of the particle in relation with time
`v=0-10t rArr v=10t` ...(iv)
This is a straight line with negative slope and passing through origin. The particle reaches fround at `t=3 s`, velocity of the particle when it hits ground is
`v=-10xx3=-30 m s^(-1)`
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For particle C: `u=0. a=-10 m s^(-2)`
Using `y=ut+(1)/(2)at^(2)`
`=0-(1)/(2)(-10)t^(2)=-5t^(2)`
It is a parabola open down and it passes through origin.
For velocity-time relation `vec v=vec u+vec at`
`v=0-10 t =-10t` ..(vi)
Straight line with negative slope passes through origion.
It will reach round at `y=-75 m`
`-75=-5t^(2) rArr t=sqrt15 s`
Velocity at the time when it reaches ground `v=-10sqrt15 s`.
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