A person jumps from the roof of a house 3.8-m high When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.73 m Part A Part B If the mass of his forso (excluding legs) is 42 kg , find the magnitude of the average force exerted on his torso by his legs during deceleration Express your answer using two significant figures. F=

To find the magnitude of the average force exerted on the torso by the legs during deceleration, we can use the equation of motion: where:- is the final velocity (0 m/s since he comes to rest),- is the initial velocity,- is the acceleration (deceleration in this case),- is the distance over which the deceleration occurs.First, we need to find the initial velocity when he starts to decelerate. We can use the equation: Rearranging for : Given:- m/s (final velocity),- (acceleration due to gravity, negative because it's deceleration),- m (distance of deceleration).Substituting the values: Now, we can find the magnitude of the average force exerted on the torso by the legs during deceleration using the equation: where:- (mass of the torso),- (acceleration due to gravity, negative because it's deceleration).Substituting the values: Therefore, the magnitude of the average force exerted on the torso by the legs during deceleration is approximately (rounded to two significant figures).