Projectile motion question?

WARNING: Long answer!

We're asked to find the range of initial speeds that a player can make a shot, with some given measurements.

Let's take a look at our known quantities:

• initial height #y_0# is #2.10# #"m"#

• final height #y# is #2.60# #"m"#

• angle of projection #alpha_0# is #38.0^"o"#

• the minimum horizontal range #Deltax_"min"# is #11.00# #"m"# #- 0.22# #"m"# #= 10.78# #"m"#

• the maximum horizontal range #Deltax_"max"# is #11.00# #"m"# #+ 0.22# #"m"# #= 11.22# #"m"#

What we can do to solve this problem is equate two different equations that say when the ball is at a height #2.60# #"m"# and a horizontal range #10.78# #"m"#.

Using the followming equations, let's solve them for #t# to eliminate them:

• #x = x_0 + (v_0cosalpha)t#

#t = (Deltax)/(v_0cosalpha) = color(purple)((10.78)/(v_0(cos38.0^"o"))#

• #y = y_0 + (v_0sinalpha)t - 1/2g t^2#

#t = (v_0sinalpha +-sqrt((-v_0sinalpha)^2 - 4(1/2g)(y-y_0)))/(2(1/2g))#

#= color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)#

Equating these two equations gives

#color(purple)((10.78)/(v_0(cos38.0^"o")) = color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)#

After solving for #v_0#, you should get a value of

#color(red)(10.76# #color(red)("m/s"#

Now let's find the maximum speed, by changing te horizontal range to #11.22# #"m"#:

#color(purple)((11.22)/(v_0(cos38.0^"o")) = color(green)((v_0sin38.0^"o" +-sqrt((-v_0sin38.0^"o")^2 - 4(4.9)(2.60-2.10)))/(9.8)#

Solving again for #v_0# gives

#color(blue)(10.96# #color(blue)("m/s"#

Therefore, the range of initial speeds is, with three significant figures,

#v_0 in [color(red)(10.8)color(white)(l)color(red)("m/s"), color(blue)(11.0)color(white)(l)color(blue)("m/s")]#

Modelling the basketball as a point projectile and ignoring air resistance.

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with a=0).

The projectile will travel a distance #S# in time #T#, where #S=11+-0.22m#, ie a minimum distance #s_1=10.78# and a maximum distance #s_2=11.22#

So we can calculate #s# using #s=ut#

# S=ucos(38)T => T=S/(ucos38)#

Vertical Motion

The projectile travels under constant acceleration due to gravity, applied vertically upwards. The projectile must travel a distance #2.6-2.1 = 0.5 \ m# vertically.

# { (s=,0.5,m),(u=,u sin(38),ms^-1),(v=,"Not Required",ms^-1),(a=,-g,ms^-2),(t=,T=S/(ucos(38)),s) :} #

Applying #s=ut+1/2at^2# we have:

# 0.5 = u sin(38) (S)/(ucos(38)) + 1/2(-g)(S/(ucos(38)))^2 #

# :. 0.5 = S tan(38) -(gS^2)/(2u^2cos^2(38)) #

And re-arranging we get:

# (gS^2)/(2u^2cos^2(38)) = S tan(38) - 0.5#
# u^2 = (gS^2)/(2cos^2(38) (S tan(38) - 0.5))#

If we take #g=9.81#, then:

With #S=S_1=10.78# we get:

# u^2 = 115.867918 ... => u=10.76419 ... #

With #S=S_2=11.22# we get:

# u^2 = 120.29947 ... => u=10.96811 ... #

As we used #g# to #2# decimal places can only expect our solution to be accurate to #2# decimal places, thus we have a rage of #u# from #u_(min) = 10.76# to #u_(max)=10.97#