Explanation:
Given that,
Initial velocity, u = 11.3 m/s
Angle above the horizontal, [tex]\theta=35^{\circ}[/tex]
Time of flight :
[tex]t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s[/tex]
Horizontal distance traveled is given by :
x = ut
x = 11.3 m/s × 1.32 s
x = 14.916 m
Maximum height is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m[/tex]
Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.
