A girl uses a slightshot to shoot a stone into the air from the roof of her school. 12m above the ground level. The...

Question

A girl uses a slightshot to shoot a stone into the air from the roof of her school. 12m above the ground level. The stone leaves the slingshot with a speed of 32m/s (ignore air friction)

a) What is the maximum height the stone can reach?

b) If the stone is fired at some angle so as to reach a max height of 30m above the point of release, what minimum speed does it achieve?

c) What is the maximum speed that the stone reaches during its flight, until striking the ground? Does this answer depend on the angle that the stone is fired at?

 

Answers: a) 64.2m, b) 20.9m/s, c) 35.5m/s

Answer 1

The girl is 12 m above ground and the ball reached a maximum height when it stops going up turns around and drops to the ground

 

                                                     max height

                                                            |

                                                            |

                                                            |  y

                                                            |

                                                            |

                                                            |

                                                            |

---------------------- roof top------------------                                 

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| y₀ = 12.0 m

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---------------------------------------------------------- ground

 

(a) h = y + y_0,

     the initial velocity of the stone is  v_0y = 32 m/s,

     at the maximum height v_y = 0 m/s,

     the acceleration due to gravity is  g= -9.8m/s², a_x = 0,

    

     use the equation: 
     

     v_y² = v_0y² +2gy ,

 

    Inserting all the values in the equation above

 

   (0)² = (32m/s)² +(2)(-9.8m/s²)(y)

 

      0  = 1024 -19.6(y)

 

      y  = 1024/19.6

 

      

       y = 52.2 m    (this is above the roof)

 

remember the girl is 12 m above the ground, therefore

 

h= y + y₀

 

y  = 52.2 + 12 = 64.2 m

 

(b) This time we need to find the value of v_y just before the stone started to descend

           

       v_0y = 32 m/s , y = 30 m

    
        v_y² = v_0y² +2gy 

 

             = (32)² + 2 (-9.8)(30)

  

             = 1024-588 = 436

 

         v_y = sqrt(436)

 

         v_y = 20.9 m/s

 

 

       v_y² = v_0y² +2a_yy  ,
     v_y² = v_0y² +2a_yy  ,