1. How much force must be applied by a soccer player to give a stationary 0.45...

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1. How much force must be applied by a soccer player to give a stationary 0.45 kg ball an acceleration of 20m/s and why? A 250 gm ice-cube is thrown in an ice-rink and it travels 40 meters. The same cube is thrown with the same horizontal velocity on a football field and it travels 5 meters. Why is that? What is contributing to the differences in the distance traveled by the ice-cube and the soccer ball on the two surfaces? sode lalim ne iw motisiq vel901 Or ort mot eqorb nevib olqmy O nA A of the ball at release? bigdor ted 2. A sprinter had an instantaneous resultant force of 3200 N acting through the heel during heel strike. If the angle of the resultant force was 125 degrees with respect to the horizontal, what were the horizontal and vertical forces acting on the heel? Estimate his body mass (in kilograms) if you know that in general vertical ground reaction force found in running are about 4.8 times the value of body weight. 1. How much force must be applied by a soccer player to give a stationary 0.45 kg ball an acceleration of 20m/s and why? A 250 gm ice-cube is thrown in an ice-rink and it travels 40 meters. The same cube is thrown with the same horizontal velocity on a football field and it travels 5 meters. Why is that? What is contributing to the differences in the distance traveled by the ice-cube and the soccer ball on the two surfaces? sode lalim ne iw motisiq vel901 Or ort mot eqorb nevib olqmy O nA A of the ball at release? bigdor ted 2. A sprinter had an instantaneous resultant force of 3200 N acting through the heel during heel strike. If the angle of the resultant force was 125 degrees with respect to the horizontal, what were the horizontal and vertical forces acting on the heel? Estimate his body mass (in kilograms) if you know that in general vertical ground reaction force found in running are about 4.8 times the value of body weight.

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1 To calculate the force required to accelerate a 045 kg ball at 20 ms2 we can use the formula Force ... View the full answer