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Nesterboy [21]
3 years ago
10

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

A) Determine the maximum height reached by the projectile(D) Determine the speed of the projectile 1.50 s after firing..
Physics
1 answer:
Margarita [4]3 years ago
6 0

Answer:

A) h = 69.58 m

D) v = 58.12 \frac{m}{s}  (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):

V_{ox} =  V_{o}Cos\alpha _{o} Formula (1)

V_{oy} =  V_{o}Sin\alpha _{o} Formula (2)

Where:

Vo: Initial velocity in m/s

\alpha_{o}: Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

v_{y}^{2} = v_{oy}^{2} -2gy Formula (3)

Where:

v_{f}^{2}: Final speed component in vertical direction in m/s

v_{oy}^{2}: 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:

v_{x} = v_{ox} Because the movement is uniform in the x direction (constant speed)

v_{y} = v_{oy}-g*t Formula (4)  Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

v_{o} = 65.2 \frac{m}{s}

\alpha _{o} = 34.5º above the horizontal

g=9.8 \frac{m}{s^{2}}

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

v_{ox} =  65.2*Cos34.5 = 53.7\frac{m}{s}

v_{oy} =  65.2*Sin34.5 = 36.93\frac{m}{s}

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):

0=(36.93)^2-2*9.8*h

h=\frac{ (36.93)^2}{2*9.8} = 69.58 m

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}

v_{x} = v_{ox} = 53.7 \frac{m}{s}

v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}  (Speed magnitude)

\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) =  tan^{-1} (\frac{22.23}{53.7})

α = 22.49° (Speed direction above the horizontal)

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