A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at an angle 55º above the horizontal. What is the impulse delivered by the foot (magnitude and direction)? The mass of the ball is .41 kg.

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Denote the impulse of the ball right before a foot strikes it as `I_1.` It is a vector directed downwards with the magnitude of `m*V_1.` We'll find the speed `V_1` later.

The resulting impulse `I_2` (a vector also) has a magnitude `m*V_2` where `V_2=18 m/s` is given. Its direction is also given.

Therefore the impulse `I_f` delivered by a foot may be found from the equation `I_1+I_f=I_2,` i.e. `I_f=I_2-I_1.`

First find the speed `V_1.` The height of a free fall is given by `h=h_0-(g t^2)/2,` where `h_0=1m,` and the speed is `V=g t.` `h=0` at `t_1=sqrt((2h_0)/g),` so

`V_1=g*t_1=sqrt(2h_0g) approx 4.4 (m/s).`

Now decompose the vectors `I_1` and `I_2` by horizontal and vertical parts.

`I_1 = (0, -m*V_1) approx (0, -1.8),`

`I_2=(m*18*cos(55°), m*18*sin(55°)) approx (4.2, 6.0).`

Therefore `I_f=I_2-I_1 approx (4.2,7.8).` The magnitude of this vector is approx. `8.9` and the angle above horizontal is `arctan(7.8/4.2) approx 62` °. This is the answer.

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