Q1. A stone is thrown vertically upwards and its reaches back to ground after 5 seconds find u and the max height stone will attain.

Q1. A stone is thrown vertically upwards and its reaches back to ground after 5 seconds find 'u' and the max height stone will attain.

Solution:

s = ut + 1/2 at^2

0 = u(5) + 1/2(-9.81)(5)^2

(9.81/2)(5×5/5) = u

u = 4.905×5

u= 24.55 m/sec

Now we find maxi height

V^2 = u^2 + 2as

0 = (24.55)^2 + 2(-9.81)s

24.55×24.55/2(9.81) = s

s = 30.72 m

Comments

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