V = 40 m/sec and theta = 34
instead of (theta), i will write (O)
the projectile will come back to the ground at point (a, 0)
so,
y = V (sin O) * t + 0.5 * g * t^2
g = -9.8 m/sec^2
so,
0 = 40 sin 34 * t - 0.5 * 9.8 * t^2
0.5 * 9.8 * t^2 - 40 sin 34 * t = 0
4.9 * t^2 - 22.37 t = 0
solve for t :
take t as a common so, t ( 4.9 t - 22.37) = 0
t = 0 OR 4.9 t - 22.37 = 0
t = 22.37/4.9 = 4.565 sec
so, the projectile will come back to the ground again after 4.565 sec
to find the point at which the projectile come to the ground
x = V (cos O ) * t = 40 * cos34 * 4.565 = 151.38 m
