Question

a particle is projected vertically upward with initial velocity of m/find the displacement and the distance covered by the particle on 6 seconds

Answer 1

$s_{n} \: = u - \frac{g}{2} (2n - 1)$
$s_{6} = m - 5(11) \\ s_{6} = m - 55$
At t= 6 u = m s= m- 55 g= -10
s = ut + 1/2at^2
m-55 = 6m - 5(36)
5m= 180 -55
m = 125/5
m = 25
Displacement = - 30 m
s= 25-55= -30 m
From this we can infer that the particle has already covered the maximum height and has proceeded 30m downwards thus we have to find the maximum height where v = 0
${v}^{2} = {u}^{2} + 2as \\ 0 = 625 - 20s \\ - 625 = - 20s \\ s = 625 \div 20 = 31.25$
maximum height is 31.25 m
hence total distance is 31.25+ 30m = 61.25m
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