Answer:
(a) The range of the projectile is 37,336.3 m
(b) The maximum height of the projectile is 5,389.03 m
Explanation:
Given;
initial velocity of the projectile, u = 650 m/s
angle of projection, θ = 30⁰
(a) The range of the projectile is calculated as;
[tex]R = \frac{u^2sin 2\theta}{g} \\\\R = \frac{(650^2)Sin (2\times 30^0)}{9.8} \\\\R = 37,336.3 \ m[/tex]
(b) The maximum height of the projectile is calculated as;
[tex]H = \frac{u^2sin^2\theta}{2g} \\\\H = \frac{(650^2)(Sin \ 30^0)^2}{2\times 9.8} \\\\H = \frac{(650^2)(0.5)^2}{19.6} \\\\H = 5,389.03 \ m[/tex]
