Question

A projectile of mass 0.850 kg is shot straight up with an initial speed of 30.0 m/s. (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only 36.7 m, determine the magnitude of the average force due to air resistance.

Answer 1

a) 45.87 m

b) 2.08 N

Step-by-step explanation:

Mass of projectile=0.85 kg=m

Velocity of projectile=30 m/s=u = initial velocity

Final velocity =0

g= acceleration due to gravity=9.81 m/s²

h=maximum height of the projectile

a) In this case loss of Kinetic energy (K.E.) = loss in potential energy (P.E.)

ΔK.E.=ΔP.E.

`(1)/(2)mu^2-(1)/(2)mv^2=mgh`

`(1)/(2)m* u^2=mgh\\\Rightarrow h=\frac {u^2}{2* g}\\\Rightarrow h=\frac {30^2}{2* 9.81}\\\therefore h=45.87\ m`

b) h'=height of the projectile=36.7 m

F=Average force due to air resistance

There will be a loss of P.E. due to air resistance

ΔP.E.=mg(h-h')

F×h'=mg(h-h')

F×36.7=0.85×9.81(45.87-36.7)

`F=(0.85* 9.81(45.87-36.7))/(36.7)`

∴ F=2.08 Newton