The Problem on Projectile Motion for AP Physics 1 and AP Physics C students

Question

The rocket is fired with a muzzle velocity of 98 m/s at an angle of 52°. After firing, the rocket accelerates with a = 28 m / s² within 5 s and moves along a straight line. The rest of the flight the rocket follows the path of the regular projectile.

Find:

A. The maximum height that the rocket has reached.

B. The range of the projectile with respect to the ground.

C. The magnitude and direction of the velocity of the rocket, when it hits the ground.

                                      

Answer 1

At first the rocket moves along a straight line with acceleration a = 28 m/s².

In the end it will have velocity

                                                          v_i = v₀ + at

After calculations we get v_i = 238 m/s.

During this time it travels a distance                           

                                                             d = (v_i² – v₀²) / 2a

Plug in the numbers and get

d = 840 m.

Hence, the height h₁ that the rocket achieved during first 5 s is

h₁ = d·sin θ or 662 m.

The total maximum height 

H = h₁ + h_max , (*)

here h_max is the height reached by the rocket during the parabolic part of flight.

The total range of the rocket

R = d₁ + d₂ , (**)

where

d₁ - horizontal distance traveled by the rocket within 5 s,

d₂ - horizontal distance covered by the rocket, when it is in parabolic motion.

d₁ = d·cos Θ or 517 m

We can find the maximum height while the rocket moves parabolically, as usually in case of projectile motion

                                  h_max = v_i²∙sin² θ / 2g

After substituting numbers, we obtain

h_max = 1759 m

Then total maximum height                     

   H = 2,421 m

Now we select x-axis to the right and y-axis vertically up. The origin of the coordinate system is the point, where the rocket starts projectile motion.

At any given time

                                             v_x = v_ix + g_x∙t

                                              v_y = v_iy + g_y∙t

Here

        v_x  - component of vector v in x direction,

        v_ix – component of initial velocity v_i in x direction,

        g_x – component of vector g in x direction

        v_y  - component of vector v in y direction,

      v_iy - component of initial velocity v_i in y direction

        g_y – component of vector g in y direction

We see that

                            v_ix = v_i∙cos θ          g_x = 0

                             v_iy  = v_i∙sin θ         g_y = - g

Finally,

                                              v_x = v_i∙cos θ                                        (1)

                                              v_y = v_i∙sin θ – gt                                (2)

Similarly for coordinates

                                                 x = v_i∙cos θ∙t                                     (3)

                                                 y = v_i∙sin θ∙t – ½ g∙t²                (4)  

When the rocket hits the ground

y = -h₁ = - 662 m.

Now I recommend to put in equation (4) numerical value to get the quadratic equation

– 662 = 188t – 5t²

"5" near t² is half of g, which we take 10 m/s², because in this problem all numbers given as whole numbers.

When we solve the quadratic equation

                                                         5t² – 188t – 662 = 0

we will get time of flight

t = 41 s

Putting in everything in equation (3), we will have

d₂ = 6,008 m

Then

R = 6,525 m

When the rocket hits the ground, its velocity vector is tangent to the parabola at this point. The magnitude of velocity

v = √ (v_x² + v_y² )

We will find components using equations (1) and (2)

Then v = 266 m/s

The vector of velocity is at angle α with horizontal and

tan α = v_y / v_x or tan α = - 1.52



So, the vector of velocity forms with the horizontal angle α = - 57º . You see that this angle is in the IV quadrant.

Answer 2

≅The motion of the rocket consists of two parts, the linear motion and the projectile motion. Note that the rocket accelerates from its initial velocity for 5 seconds.

Thus v=98+5(28)=98+140=238m/s

We also need the height traveled during this time. We use the equation for straight line motion and then multiply the value by sin52^o since we need the y component.

s=v_ot+0.5at²=(98)(5)+0.5(28)(5)²=490+350=840m

Thus x=840 cos 52^o ≈ 517 and y= 840 sin 52^o ≈ 662 m when the projectile motion begins.

Remember the velocity of the projctile is constant (238 cos 52^o) in the x direction while the y component (238 sin 52^o) is affected by the acceleration due to gravity. The projectile motion ends when the rocket returns to ground level.

The maximum height reached is the 840 sin 52^o + (238²sin²52^o)/2(9.8)

determined by the height before projectile motion starts plus the maximum height of projectile (v_y²/2g)

The equation for the projectile motion for the x direction is x_p= (238 cos 52^o)t

and the equation in the y direction is y_p= 840 sin 52^o+ (238 sin 52^o)t-0.5(9.8)t²

which we set to zero to determine when the rocket hits the ground. So we will solve for t using the quadratic formula.

t= [-238 sin 52^o+√(238²sin²52^o)-4(840sin 25^o)(4.9)]/9.8

c. The y component of the velocity when the rocket hits the ground is v_y=(238 sin 52^o)-9.8t

using the t value from the previous part of the problem whlie the v_x= 238 cos 52^o

v=√v_x²+v_y² and the direction is given by θ= arctan(v_y/v_x)