Question

Closed

Answer 1

30 m/s
discussion question for when u r online

Answer 2

i got the answer as 29.56m/s. here's how.
first, the situation is something like this:

let u = velocity with which ball is thrown
v1 = velocity at pt where it crosses the 25m mark.
v2 = velocity at the highest pt = 0m/s
first, use the eq v^2 = u^2 + 2as to get a relation between u and v1.
(v1)^2 = u^2 - 2g(25)
(v1)^2 = u^2 -50g..................(i)
now, use the eq v = u + at taking the starting pt as the instant when the ball first crosses the 25m mark. consider only its journey till the highest pt. so time taken to cover this distance will be 4/2 = 2sec. now substitute.
v2 = v1 - 2g
0 = v1 - 2g
v1 = 2g
substitute this value in (i)
4g^2 = u^2 - 50g
sub g = 9.8m/s^2
and u'll get u = 29.56m/s