vyi = 37.0 sin52,8° = 29.47 m/s
vix = 37.0 cos52.8° = 22.37 m/s
(a) the ball's height y is
y(t) = 29.47t - 4.9t^2
So, when it hits the green,
29.47t - 4.9t^2 = 5.3
t = 5.829
(b) the horizontal distance, at constant speed is
x = 5.829 * 22.37 = 130.39
(c) the vertex of the parabola is at
t = -b/2a = 29.47/9.8 = 3.01
y(3.007) = 44.31
A golfer hits his approach shot at an angle of
θ = 52.8°,
giving the ball an initial speed of
v0 = 37.0 m/s
(see figure below). The ball lands on the elevated green,
yf = 5.3 m
above the initial position near the hole, and stops immediately.
(a) How much time passed while the ball was in the air?
s
(b) How far did the ball travel horizontally before landing?
m
(c) What was the peak height reached by the ball?
m
1 answer
1 answer