Question

A body is projected with a spped of 40m/s vertically up from the ground . what is the maximum height reached by the body?what is the entire time of motion?what is the velocity at 5 seconds after the projection?take g=10 m/s

Answer 1

Sol. u = 50 m/s, g = –10 m/s2 when moving upward, v = 0 (at highest point). a) S = (V^2-u^2)/2a = (0-〖50〗^2)/(2(-10)) = 125 m maximum height reached = 125 m b) t = (v – u)/a = (0 – 50)/–10 = 5 sec c) s’ = 125/2 = 62.5 m, u = 50 m/s, a = –10 m/s2, v2 – u2 = 2as ⇒ v = √(( u^2+2as)) = √(〖50〗^2+2(-10)(62.5) ) = 35 m/s.
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Answer 2

Answer:

Radius of sphere=2cm =2/100=0.02 m

Volume of the sphere=4/3

.

r³

=4/3× 22/7×0.02×0.02m³=0.000704/21 m³

Mass of sphere=0.05kg.

Relative density of sphere=density ×10³

=0.05/0.000704\21 ×10

=0.05×21/704×10-6×10³

=1.05×10³/704×10-6

=1.05×10³±6/704=1.05×109/704=1.49×106

Relative density of sphere= 1.49×106 kg/