Question

projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The projectile hits the incline plane at point M. 
a) Find the time it takes for the projectile to hit the incline plane.
b)Find the distance OM. ​

Answer 1

Answer:

Transcribed Image Text from this Question

A projectile is launched from point O at an angle 22 degree with an initial velocity of 15 m/s up an incline plane that makes an angle of 10 degree with the horizontal. The projectile hits the incline plane at point M. Find the time it takes for the projectile to hit the incline plane. The 2-kg collar is released from rest at A and along the smooth vertical guide. The spring has an unstretched length of 200 mm and a stiffness k of 600 N/m. Determine: (a) The speed of the collar when it reaches position B. (b) The normal force exerted on the collar at this position.

Answer 2

Given:

• a projectile launched from point O at an angle of 22°
• initial velocity 15m/sec.
• incline plane makes an angle with the horizontal of 10°
• The projectile hits the inclined plane at point M

To find:

• the time it takes for the projectile to hit the inclined plane.
• the distance OM

explanation:

the formula of the components of the displacement in the case of a projectile.

The x component of the displacement is given as,

                                         $x = V_{0}$cos(Ф)t

Where $V_{0}$ is the velocity of the motion, t is the time taken and Ф is the angle made by the projectile.

The y component of the displacement is given as,

                                        $y = V_{0}$ sin(Ф)t $- (\frac{1}{2})gt^2$

                                (g is the gravitational constant)

The angle made by the projectile,

                                        Ф = 22+10 = 32°

The velocity of the motion, $V_{0}$ = 15m/s

The relationship between the coordinates x and y. with the inclination is given as follows,

                                         tan10° = $\frac{y}{x}$

let us substitute the given values,

                                          tan10° = {$V_{0}$ sin(Ф)$-(\frac{1}{2})gt^2$} ÷ {$V_{0}$ cos(Ф)$t$}

                                           $(\frac{1}{2} )gt + V_{0}$ cos(Ф)tan10°$-V_{0}$ sin(Ф) = 0

                                              $t$ = {$V_{0}$sin(Ф)$-V_{0}$cos(Ф) tan(Ф)} ÷ {0.5g}

                                              $t$ = {$V_{0}$ sin32°$-V_{0}$ cos32° tan10°} ÷ {0.5×9.8}

                                              $t = 1.16 sec$

Therefore, the time taken is 1.16 s.

Now, we will find the distance OM.

From the figure, it's clear that the distance OM is given as follows,

             = OM ÷ $\sqrt{}${($V_{0}$ cos (Ф)$t^2$ + ($V_{0}$ sin(Ф)t $-(\frac{1}{2} )gt)^2$}

             = OM ÷ $\sqrt{}${( 15×cos32° × $1.16)^2$ + ( 15×sin32°× 1.16$-(\frac{1}{2})$×9.8×1.16)}

             OM = 15m

Hence the value of the distance OM is 15 m and time taken is 1.16sec.