A football is kicked at an angle of 45 degrees from a distance of 34 meters and just clears a 3 meter high goal on its way down. To the nearest m/s what was its initial speed?
1 Answer
Explanation:
WARNING: Rigorous Solution!
We're asked to find the initial speed of a projectile (football) given it was launched at an angle of
Here's what we know:
If it just clears the goal, with height of
and
We can use the kinematics equations
and
(I understand this is Precalculus, and you may want an answer involving parametric equations, but here's a Physics approach! These are equations used when an object is treated as a projectile, so don't worry now where these equations come from.)
Parametric equations will not give you the exact answer like this solution will (as terrifying as it is; most of it is just solving for a variable).
where

#Deltay# is the change in#y# position, in this case#3# #"m"# 
#Deltax# is the change in#x# position, given as#34# #"m"# 
#v_0# is the initial speed (what we're trying to find) 
#alpha_0# is the launch angle, given as#45^"o"# 
#t# is the time (we'll use this variable, even though time is not given) 
#g# is the acceleration due to gravity near earth's surface, equal to#9.81# #"m/s"^2#
We recognize that the projectile has an
Therefore, what we can do is isolate
Isolating
Isolating
Since
(
I eliminated the
Here's solving for
Cross multiply:
#v_0(v_0sinalpha_0 + sqrt((v_0sinalpha)^2  2g(Deltay))) = (g)(Deltax)(secalpha_0)#
##
#v_0(v_0sinalpha_0 + sqrt((v_0)^2sin^2alpha  2g(Deltay))) = (g)(Deltax)(secalpha_0)#
##
#(v_0)^2sinalpha_0 + v_0sqrt((v_0)^2sin^2alpha  2g(Deltay)) = (g)(Deltax)(secalpha_0)#
##
#v_0sqrt((v_0)^2sin^2alpha  2g(Deltay)) = (g)(Deltax)(secalpha_0)  (v_0)^2sinalpha_0#
##
#(v_0)^2((v_0)^2sin^2alpha_0  2g(Deltay)) = ((g)(Deltax)(secalpha_0)  (v_0)^2sinalpha_0)^2#
##
#(v_0)^4sin^2alpha_0  (v_0)^2 2g(Deltay) = ((g)(Deltax)(secalpha_0)  (v_0)^2sinalpha_0)^2#
##
#(v_0)^4sin^2alpha_0  (v_0)^2 2g(Deltay) = g^2(Deltax)^2sec^2alpha_0 + (v_0)^4sin^2alpha_0 2g(Deltax)(v_0)^2 tanalpha_0#
##
#2g(Deltay)(v_0)^2  g^2(Deltax)^2sec^2alpha_0 + 2g(Deltax)(v_0)^2 tanalpha_0 = 0#
##
Collect in terms of#v_0# :
#(v_0)^2(2g(Deltax)tanalpha_0  2g(Deltay))  g^2(Deltax)^2sec^2alpha_0 = 0#
##
#(v_0)^2(2g(Deltax)tanalpha_0  2g(Deltay)) = g^2(Deltax)^2sec^2alpha_0#
##
Divide both sides by#2g(Deltax)tanalpha_0  2g(Deltay)# :
#(v_0)^2 = (g^2(Deltax)^2sec^2alpha_0)/(2g(Deltax)tanalpha_0  2g(Deltay))#
##
#color(blue)(v_0 = (sqrt(g) (Deltax)secalpha_0)/(sqrt2 sqrt((Deltax)tanalpha_0  Deltay))#
In all honesty, you could have just plugged in values from the start and solved for
Plugging in our known values, we have
It's clearly beautiful:)
We can even check this to make sure our solution is correct, by plugging everything into our original equation:
(slight rounding errors led to this, but it is correct!)