Physics Problem Help Please! Newtons Laws?
Hello, I am new here and am hoping someone can help me understand how to work this problem.
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.290 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.980 m above the floor.
(a)Calculate his velocity (in m/s) when he leaves the floor.
(b)Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.290 m.
(c)Calculate the force (in N) he exerts on the floor to do this, given that his mass is 108 kg.
Hello, I am new here and am hoping someone can help me understand how to work this problem.
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.290 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.980 m above the floor.
(a)Calculate his velocity (in m/s) when he leaves the floor.
(b)Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.290 m.
(c)Calculate the force (in N) he exerts on the floor to do this, given that his mass is 108 kg.
1 Answer
As the basketball player leaves the ground, his motion is like a projectile. Applicable kinematic expression is
#v^2-u^2=2as#
(a) When he reaches maximum height of
Noting that acceleration due to gravity acts in a direction which is against the initial direction of motion, hence it is taken as
#0^2-u^2=2(-9.81)xx0.980#
#=>u=sqrt(2(9.81)xx0.980)#
#=>u=4.385\ ms^-1#
(b) His acceleration when he straightens his legs is found from same kinematic expression. Inserting given values in this case we get
#(4.385)^2-(0)^2=2axx0.290#
#=>a=(4.385)^2/(2xx0.290)#
#=>a=33.151\ ms^-2#
(c) We see that net force
As follows from Newtons' Third Law of motion that action and reaction are equal and opposite, we see that Force applied on the floor must be Sum of Net upwards force and downwards weight of the player,
#F_"Total"=F_"net"+mg#
#=>F_"Total"=ma+mg=m(a+g)#
Inserting various values we get
