Question

A cannon fires an acrobat into the air at an
angle of 45
above the horizontal such that she
reaches a height h above her launch height. If the
cannon is now aimed straight upwards, what is
the maximum height reached by the same
acrobat

Answer 1

first u need to know the formula
h(maximum ht above ground)=(usin@)^2/2g
so apply that with g=10m/s^2 h and angle 45
then u get initial velocity u in terms of h
when they say u aim verticallly it means straight up with 90 degree angle with vertical
take v=0
use v^2-u^2=2as
find s ht it reaches
u have any doubts on this problem u can ask me

Answer 2

oh man... accidently deleted the question about the speed trap, real fast, how do you solve for t?
10t = t^2?