At what angle must a projectile be fired to travel 3.624m from a height of 1.012m?

x=3.624m
y=1.012m
v0=?
t=?
v=?

1 Answer
Sep 16, 2017

Here it is a approach of formula...
We know for the projectile motion the following formulae,
#H=(U^2sin^2theta)/2g# , #R=(U^2sin2theta)/g# , #t=(2Usintheta)/g#
Here #H# is the height of the projectile along #Y# axis and #R# is the range along #X# axis. #U# is the initial velocity and # theta# is the angle of projectile thrown.
Here we get , #H=1.012m# and #R=3.624m#

Answer is incomplete.....
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