A student of mass m pushes off from the floor with an average force F exerted for a time (delta)t.

(a) Using impulse and energy ideas, derive a symbolic expression for the maximum height (y) the student jumps in the air.

(b) Solve for this jump height when a 60.0-kg student pushes against the floor with a force of 1111 N for 0.220 s.

 

Answer:

a) ?h=F2Δt2??/2gm2

b) ?h=84.7cm?

Step-by-step explanation

Given :

Mass of the student, ?m=60kg?

Force applied by the student on the floor, ?F=1111N?

Time for which the force is applied, ?Δt=0.22s?

Acceleration due to gravity, ?g=9.8m/s2?

Initial velocity of the student, ?u=0m/s?

To Find :

Maximum height that the student jumps, ?h?

Solution :

a) Applying the impulse on the student for the time of application of force, we have

Impulse ?=? Change in momentum

?⇒I=ΔP?

?⇒FΔt=Pfinal?−Pinitial??

?⇒FΔt=mv−mu?

?⇒FΔt=mv−m×0?

?⇒FΔt=mv?

?⇒v=mFΔt??

From the work energy theorem, we have

?Wgravity?=ΔKE?

?⇒mg(−h)=21?m(02−v2)?

?⇒mgh=21?mv2?

?⇒2gh=v2?

?⇒h=2gv2??

?⇒h=2gm2F2Δt2??

b)

Putting in the numerical values, we get

?h=2×9.8×60211112×0.222??

?⇒h=7056059741.14??

?⇒h=0.847m?

?⇒h=84.7cm?

PFA