A golfer hits a shot to a green that is elevated 3.30 m above the point where the ball is struck. The ball leaves the club at a speed of 19.0 m/s at an angle of 52.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands. (Write on paper)
horizontal component of velocity :
velocity vx = vcostheta
= (19.0 m/s)cos52
= 11.69 m/s
vertical component of the velocity :
velocity vy = vsintheta
= (19.0 m/s)sin52
= 14.97 m/s
but , at maximum height vy = 0 m/s
from kinematic equations ,
v2 = v0y2 + 2as
(0 m/s) = v0y2 + 2ay
(0 m/s) = v0y2 + 2(-g)y
y = (v0y)2/2g
= (14.97 m/s)2/2(9.8 m/s2)
= 11.43 m
therefore , Δy = 11.43 m - 3.30 m
= 8.13 m
from kinematic equations,
vfy2 = v0y2 + 2as
vfy2 = (0 m/s)2 + 2g(Δy)
vfy2 = 2(9.8 m/s2)(8.13 m)
vfy = 12.623 m/s
therefore , resultant velocity is
v = square root [(11.69 m/s)2 + (12.623 m/s)2]
= 17.20 m/s Answer
I hope help you !!
A golfer hits a shot to a green that is elevated 3.30 m above the point...
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