A golf ball with an initial angle of 32° lands exactly 240 m down the range on a level course.?

Question

(a) Neglecting air friction, what initial speed would achieve this result?


(b) Using the speed determined in item (a), find the maximum height reached by the ball.

Answer 1

x(t)=v0*cos(32)*t
240=v0*cos(32)*t
v0=240/(cos(32)*t)

y(t)=v0*sin(32)*t-.5*g*t^2
0=240*tan(32)-5*g*t^2
solve for t

t=5.53 s
v0=51.18

max height is found from
vy(t)=v0*sin(32)*t-g*t
the t at max height is 1/2 the flight time
y(5.53/2)=37.5 m

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