You throw a ball vertically upword. The ball leaves your horn with a speed of 15

Question

You throw a ball vertically upword. The ball leaves your horn with a speed of 15.0m/c the ball is then in free foil. What is the maxima height above your hand reached by the ball? What is the boll's acceleration when it is at its maximum height? You wait and catch the ball on its way down. How much time was the ball m the air? (assume your hand .s at the same position as when you threw the ball up) If you missed the boll an its way down, find its speed just befor it the hit the ground 1.2m below your hand.

Explanation / Answer

Intial velocity u = 15.0 m/s

Accelaration due to gravity be g = 9.8 m/s2

1)Time taken by it to travel 1 second be t1 = 1 s

Now the velocity of ball V1 at time t1 is given by

                          V1 = u –( g * t1 ) = 15 – ( 9.8 * 1)

                                                  = 5.2 m/s

2) the velocity of the ball after time t2 = 2 seconds is

                           V2 = u- ( g * t2) = 15 – ( 9.8 * 2 )

                               = 15 – 19.6 = - 4.6 m/s

3) maximum height reached by the ball is H = u2/2 *g

                        H = ( 15 )2/2 * 9.8 = 225/19.6

                           = 11.47 m

4) generally the velocity of the object is zero at maximum height but it does not mean that acceleration is zero.In the case of free fall or vertical projection g is constant.But till the ball reaches the maximum height g is negative and decreases the velocity of the ball .But after ball reaches the maximum height g reverses the direction and becomes positive. So the velocity of the ball goes on increases during the fall

5)Time taken to catch the ball is equal to the time of flight of the ball

Time of flight is given by T = 2 * u/g

                                                = 2 * 15/9.8

                                                = 3.06 second

6)Velocity of the ball after reaching the point of projection Vp is equal to intial velocity

                         Vp = - u = - 15 m/s

Now height left above the ground is H1 = 1.2 m

Then Vg2 – Vp2 = 2 * g * H1 where Vg is velocity of ball at the ground

              v2 – u2 = 2* g * h

              Vg2 – ( 15 )2 = 2 * 9.8 * 1.2

              Vg2 – 225     = 23.52

                              Vg2 = 23.52 + 225 = 248.52

                               Vg = 15.76 m/s

For simplicity in calculation we can take g = 10 m/s2, then values are V1 = 5 m/s , V2 = -5 m/s

H = 11.25 m , T = 3 s , Vg = 15.78 m/s

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