the range is
R = v^2/g sin2θ
so plug and chug
now you want |R-170|
A golfer is teeing off on a 170.0 m long par 3 hole. The ball leaves with a velocity of 40.0 m/s at 50.0o to the horizontal. Assuming that she hits the ball on a direct path to the hole, how far from the hole will the ball land (no bounces or rolls)
2 answers
vertical problem:
a = -g = -9.81 m/s^2
v = Vi + a t = 40 sin 50 - 9.81 t = 30.64 - 9.81 t
when will it be at the top?
when v = 0 obviously
so at top
t = 30.64 / 9.81 = 3.12 seconds
so how long is it in the air?
Twice that long, the top is the vertex of the parabola :)
so
flight time = 6.24 seconds (wow)
THAT is all we need about up and down, now how far?
Horizontal problem
u = constant speed(no horizontal force ) = 40 cos 50 = 25.7 m/s
so 25.7 m/s for 6.24s = about 160.4 meters, less than ten meters to the hole. I think we are on the green !!
a = -g = -9.81 m/s^2
v = Vi + a t = 40 sin 50 - 9.81 t = 30.64 - 9.81 t
when will it be at the top?
when v = 0 obviously
so at top
t = 30.64 / 9.81 = 3.12 seconds
so how long is it in the air?
Twice that long, the top is the vertex of the parabola :)
so
flight time = 6.24 seconds (wow)
THAT is all we need about up and down, now how far?
Horizontal problem
u = constant speed(no horizontal force ) = 40 cos 50 = 25.7 m/s
so 25.7 m/s for 6.24s = about 160.4 meters, less than ten meters to the hole. I think we are on the green !!