A ball projected with an initial velocity u at an angle tita to the horizontal cover a horizontal range of 26m after 5sec. Cal. The value of u and tita? g=10ms-1

Answer 1

well I call initial velocity horizontal component u and vertical component Vi
but your way
horizontal velocity = z = u cos theta
initial vertical velocity = Vi = u sin theta

z = 26/5 = u cos theta = 5.2

now it tskes 5 seconds to go up and down
so it takes 2.5 seconds to go up
v = Vi - 10 t
0 = Vi - 10 t
so
t = u sin theta/10 = 2.5
so
u sin theta = 25
therefore
u sin theta/u cos thets = 25/5.2
or
tan theta = 4.81
theta = 78.3 degrees
and u = 25/sin 78.3