At the highest point of its path, a body has a velocity of 10 m/s and is 8m above the ground. Find the angle of projection and initial velocity.
x''=0
x'=c
x=ct+d
y''=-10
y'=-10t+e
y=-5t^2+et asuming launch from 'the ground'
At the highest point of its path y-velocity = y'= 0, so x'=10 m/s.
y'=0 when y=8 ----> 10t=e ----> t=e/10 ----->8=-5(e/10)^2+e^2/10 ---> e=12.649
y'(0)=e=12.649
x'(0)=x'=10
initial velocity = 16.1245 m/s
launch angle = 51.67 degrees
NB i have used g=10m/s^2.