math nahi ata
vyhef7 mm ,hvg67 d
c cv hnhgcyjmwmye47;z,l4ghghfyhn ghcy
A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.8 m/s at an angle of 53.0˚ 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.
2 answers
h(t) = 3.00 + 17.8 sin53.0˚ - 4.9t^2
find t when h=0
Then, recall that the upward speed is
vy = 17.8 sin53.0˚ - 9.8t
vx = 17.8 cos53.0˚
the final v is of course found by
v^2 = vx^2 + vy^2
find t when h=0
Then, recall that the upward speed is
vy = 17.8 sin53.0˚ - 9.8t
vx = 17.8 cos53.0˚
the final v is of course found by
v^2 = vx^2 + vy^2