Question

1) A body is projected vertically upward with velocity v. Find its velocity at half of its max. height. 2) The velocity-time relation of particle is given by v=kt, where k=2m/s².Find the distance travelled in first 3 sec.

Answer 1

Answer:

we have

H = max height = (vf^2 - v^2)/2g

H = (0 - v^2)/-10 = v^2/10

so at H/2

H/2 = (vf^2 - 0)/10

so vf = v/√2

Answer - v/√2

v = kt = 2t

ds = vdt => s = t^2

in 3 sec,

s = (3)^2 = 9m

Answer - 9 meters