Aprojectile 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..

A)

D)  (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):

Formula (1)

Formula (2)

Where:

Vo: Initial velocity in m/s

: Initial angle above the horizontal in grades

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

Formula (3)

Where:

: Final speed component in vertical direction in m/s

: 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:

Because the movement is uniform in the x direction (constant speed)

Formula (4)  Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

= 34.5º above the horizontal

Development of the problem:

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

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):

D) Speed of the projectile 1.50s after firing

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

 (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)