A golfer tees off from the top of a rise, giving the golf ball an initial veloci
ID: 1276689 • Letter: A
Question
A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 41 m/s at an angle of 25 above the horizontal. The ball strikes the fairway a horizontal distance of 190 m from the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 41 m/s at an angle of 25 above the horizontal. The ball strikes the fairway a horizontal distance of 190 m from the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?
Explanation / Answer
x direction
x = vx t
190 = 41*cos(25 degrees)*t
t = 5.11 s
y direction
y = y0 + v0y t + 1/2 a t^2
0 = h + 41*sin(25 degrees)*5.11 - 0.5*9.81*5.11^2
h=39.5 m
b)
conservation of energy
1/2 mv^2 + m g h = 1/2 mv^2
0.5*41^2 + 9.81*39.5 = 0.5*v^2
v=49.56 m/s
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