Vikas TU
Last Activity: 8 Years ago
Time of flight is actually given for crossing the point actually two times.
with projection at hetha = 90 degree
4 = 2v/g
v = 2g m/s
This is the velocity at 25 meter initially (crossing first time).
Now calculating velocity (initaial)
at height of 25 m.
.
Sign convention taking upwards positive and downwards negative.
From 3rd law of eqn.
v^2 = u^2 – 2gs
(2g)^2 = u^2 – 2g*25
u^2 = 4g^2 +50g
u^2 = 400 + 500
u^2 = 900
u = 30 m/s
Hence the velocity with the ball was thrown was 30 m/s.
Plz Approve!
