A golfer hits a golf ball which leaves the club with an initial velocity of 52m/
ID: 1653922 • Letter: A
Question
A golfer hits a golf ball which leaves the club with an initial velocity of 52m/s at an angle of 49 with respect to the horizontal. The ball travels to a max height and after some distance, on its way down hits a lamp post. The time of flight from the golf ball is hit to when it strikes the lamp post is 7.4s.
A.) Determine the horizontal and vertical components of the initial velocity of the golf ball just after being hit.
B.) Determine the time from the moment it was initially hit, that it takes the ball to recah max height.
C.) Determine the max height above the ground level reached by the ball
D.) Determine the height above the ground where the ball hits the lamp post
E.) Determine the horizontal distance from where the ball is initially hit to the base of the lamp post that it struck
F.) Determine the x and y components of the final velocity of the ball just before it strikes the lamp post
G.) Calculate the magnitude of the final velocity vector of the ball just before it strikes the lamp post and the direction angle it makes with respect to the horizontal x direction.
Explanation / Answer
A) horizontal velocity component = 52cos49 = 34.12m/s
vertical velocity component = 52sin49 =39.24m/s
B) Time needed to reach maximum height = 39.24/9.8 = 4 seconds
C)Maximum height reached = 39.242/(2*9.8) = 78.56m
D)The ball hits the lamp post after 7.4 seconds
Height by which ball falls in 7.4-4 = 3.4 seconds = (1/2)*9.8*3.42 = 56.64m
Height of lamp post = 78.56-56.64=21.92m
E) Horizontal distance = 34.12*7.4 = 252.5m
F)x component of velocity = 34.12 m/s
y component of velocity = 9.8*3.4=33.32 m/s
G)Magnitude of velocity = sqrt(34.122+33.322) = 47.69m/s
Angle it makes with horizontal x direction = tan-1(33.32/34.12) = 44.32o (measured clockwise)
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