Answer:
A) [tex]h = 69.58 m[/tex]
D) [tex]v = 58.12 \frac{m}{s}[/tex] (Speed magnitude)
α = 22.49° (Speed direction above the horizontal)
Explanation:
Conceptual analysis:
To solve this problem we consider the following concepts:
1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.
The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):
[tex]V_{ox} = V_{o}Cos\alpha _{o}[/tex] Formula (1)
[tex]V_{oy} = V_{o}Sin\alpha _{o}[/tex] Formula (2)
Where:
Vo: Initial velocity in m/s
[tex]\alpha_{o} [/tex]: Initial angle above the horizontal in grades
2) The formula to calculate its velocity at any vertical position(y) is as follows:
[tex]v_{y}^{2} = v_{oy}^{2} -2gy[/tex] Formula (3)
Where:
[tex]v_{f}^{2}[/tex]: Final speed component in vertical direction in m/s
[tex]v_{oy}^{2}[/tex]: Initial speed component in vertical direction in m/s
g: acceleration due to gravity in m/s2
3) The formulas to calculate the projectile velocity components at any time (t) are:
[tex]v_{x} = v_{ox}[/tex] Because the movement is uniform in the x direction (constant speed)
[tex]v_{y} = v_{oy}-g*t[/tex] Formula (4) Because the movement is uniformly accelerated in the y direction
Known information:
We know the following data:
[tex]v_{o} = 65.2 \frac{m}{s}[/tex]
[tex]\alpha _{o}[/tex] = 34.5º above the horizontal
[tex]g=9.8 \frac{m}{s^{2}}[/tex]
Development of the problem:
Initial speed components(Vox, Voy), (Formula (1), Formula (2)
[tex]v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}[/tex]
[tex]v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}[/tex]
A) Maximum height (h):
When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):
[tex]0=(36.93)^2-2*9.8*h[/tex]
[tex]h=\frac{ (36.93)^2}{2*9.8} = 69.58 m[/tex]
D) Speed of the projectile 1.50s after firing
We replace t=1.5 s in the formula(4)
[tex]v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s} [/tex]
[tex]v_{x} = v_{ox} = 53.7 \frac{m}{s} [/tex]
[tex]v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}[/tex] (Speed magnitude)
[tex]\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})[/tex]
α = 22.49° (Speed direction above the horizontal)
