Answer:
a
The height is [tex]H = 6.74 \ m [/tex]
b
The horizontal distance is [tex]D = 23.74 \ m [/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 15 \ m/s[/tex]
The angle is [tex]\theta = 40^o[/tex]
The height of the cannon from the ground is h = 2 m
The distance of the net from the ground is k = 1 m
Generally the maximum height she reaches is mathematically represented as
[tex]H = \frac{v^2 sin^2 \theta }{2 * g } + h[/tex]
=> [tex]H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2[/tex]
=> [tex]H = 6.74 \ m [/tex]
Generally from kinematic equation
[tex]s = ut + \frac{1}{2} at^2[/tex]
Here s is the displacement which is mathematically represented as
s = [-(h-k)]
=> s = -(2-1)
=> s = -1 m
There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half
a = -g = -9.8
[tex]u = v sin (\theta)[/tex]
So
[tex]-1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2[/tex]
=> [tex]-4.9t^2 + 9.6418t + 1 = 0[/tex]
using quadratic formula to solve the equation we have
[tex]t = 2.07 \ s[/tex]
Generally distance covered along the horizontal is
[tex]D = v cos (40) * 2.07[/tex]
=> [tex]D = 15 cos (40) * 2.07[/tex]
=> [tex]D = 23.74 \ m [/tex]
