A golf and physics question. give it a shot.?

Question

A golf ball with an initial angle of 38° lands exactly 220 m down the range on a level course.
(a) Neglecting air friction, what initial speed would achieve this result?
m/s
(b) Using the speed determined in item (a), find the maximum height reached by the ball.
m

Answer 1

a) The initial velocity will be called V
The distance traveled is 220m
without friction this is the horizontal velocity
Hv , times time of flight, t

220=Hv* t

Hv=V*cos(38)
so 220=V*cos(38)*t

Next, the average velocity to apogee
is 1/2 * m*g* t/2
(t/2 since t is total flight time)

using conservation of energy
1/2*m*g*t^2/4 is the potential energy gained to apogee

the vertical component of initial velocity is Vv
Vv=V*sin(38)
the average velocity to apogee is 1/2*V*sin(38)
and the work is 1/2 *m*V*sin(38)*t/2
which means that
1/2 *m*V*sin(38)*t/2 = 1/2*m*g*t^2/4
or
V*sin(38)=1/2*g*t
or
2*V*sin(38)/g=t
using 220=V*cos(38)*t
or
220/(V*cos(38))=t
subtract the two equations
2*V*sin(38)/g - 220/(V*cos(38)) =0
divide by 2
V*sin(38)/g - 110/(V*cos(38)) =0
multiply by V*cos(38)
V^2 * Sin(38)* cos(38)/g =110
solve for V^2
V^2= 110*g/(sin(38)*cos(38))
solve for V (the positive root)
V=47 m/s

b)
now solve for t
220/(V*cos(38))=t
t=5.9 seconds

the apogee is 1/2 * sin(38) * V * t
=86m

j