1.

A particle is thrown upwards with a spupward direction, and (b) the maximum height reached.eed of 39.2 m/s. Find (a) the time for which it moves in the

Answer»

Intial velocity(u) = 39.2 m/sfinal velocity(v) = 0

acceleration(a) = -9.8

a) using third equation of motion

2as = v^2 -u^22 (-9.8)(s)= (0)^2 - (39.2)^2 -19.6 s = - 1536.6419.6s = 1536.64s = 1536.64 / 19.6= 78.4

so,the maximum height covered = 78.4 m

b) let time taken be t

using acceleration formula

a= (v-u)/t-9.8 = (0- 39.2)/t-9.8 t = -39.29.8t = 39.2t = 39.2 / 9.8= 4

Thanks



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