rifle AA point. is aimed horizontally at a target 100 away. bullet hits the targ

Question

rifle AA point. is aimed horizontally at a target 100 away. bullet hits the target 1.50 cm below the aimin a) What is the The bullet's flight time? S t- 55.3ms oln: b) What is the muzzle speed of the bullet? Soln: vil 1810 Homework AB A rock is thrown at an initial velocity of 6.20 m/s at an angle of 65 above the horizontal. How long does it take the rock to reach a position one meter above the point of release during its ascentn Soln: t- 0.220 s Homework 40 A projectile is launched at an angle of 18.0 degrees below the horizontal at a speed l15 ml the top of a 65.0 building. of a) How far horizontally from the launch point does the projectile land if the ground at the base ofthe building is level? Soln x -x 166m b) How long does it take to reach a point 20.0m vertically the launch point solnet m 0525s below

Explanation / Answer

4A.a)

Distance travelled by bullet vertically= 1.5 cm=0.015 m

Since, Bullet is fired horizontally, initial vertical speed of the bullet=0.

We know, s= ut+ 1/2 gt2 , here s is distance travelled

                                                u is the initial speed

                                                g is acceleration due to gravity and t is time.

   So,    0.015= 0*t+ 1/2 *9.8*t2

                 t= (0.015/4.9)1/2

             t=0.0553 seconds =55.3 ms

4.A.b.) Muzzle speed of the bullet is the speed by which it leaves the muzzle.

      Muzzle speed = horizontal distance travelled/ time taken

                           = 100/0.0553=1807.4= 1810 m/s

4.B) We need to determine the time required by rock to reach a height of 1 meter after it is realesed.

      Vertical component of velocity= V sin theta= 6.20 sin 65=5.66 m/s

Now, applying s=ut+1/2gt^2

                      1= 5.66 t- 1/2*9.8*t^2

                      4.9t2-5.66t+1=0

                      t= 0.22 sec or 0.94

                Lower one will be our answer.

                t=0.22 second

4C.a) Verical component of velocity, V= 115 sin 18=35.54 m/s

         Horizontal Component of velocity= U= 115 cos 18=109.37m/s

       Let the time of flight be t.

                          65= 35.54 t+ 1/2*9.8*t^2

                         4.9t^2+35.54t-65=0

                        t=1.51 and -8.76

                  Neglecting the negative value, t=1.51

    Horizontal distance travelled= U*t=109.38*1.51=166m

b. s=20m

    20= 35.54 t+1/2*t2* 9.8

    4.9t^2+35.45t-20=0

    t=0.525 and -7.76

Neglecting the negative value, t=0.525 s

   

     

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