Answer:
9.58 m/s
Explanation:
x = 5.6 m
y = 3 m
θ = 60°
Let u be the initial velocity of projection.
The general equation of projectile path is given by
[tex]y = x tan\theta -\frac{gx^{2}}{2u^{2}Cos^{2}\theta }[/tex]
[tex]3 = 5.6\times tan60 -\frac{9.8\times 5.6 \times 5.6}{2\times u^{2}Cos^{2}60}[/tex]
[tex]\frac{614.66}{u^{2}}=6.7[/tex]
u = 9.58 m/s
thus, the velocity of projection is given by 9.58 m/s.
