A golfer, standing on a fairway, hits a shot to a green that is elevated 6.93 m

Question

A golfer, standing on a fairway, hits a shot to a green that is elevated 6.93 m above the point where she is standing. If the ball leaves her club with a velocity of 48.9 m/s at an angle of 41.5 ° above the ground, find the time that the ball is in the air before it hits the green.

2)From the top of a cliff overlooking a lake, a person throws two stones, as shown in the drawing. The cliff is 38.7 m high. The two stones described have identical initial speeds of v0 = 10.3 m/s and are thrown at an angle = 22.1 °, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the water? Neglect air resistance.

Explanation / Answer

here,

1)

theta = 48.9 degree

initial speed , u = 41.5 m/s

y = 6.93 m

let the time passed while the ball was in the air be t

y = u*sin(theta) * t - 0.5 * g * t^2

6.93 = 41.5* sin(48.9) * t - 0.5 * 9.81 * t^2

solving for t

t = 0.22 s or 6.14 s

for landing of ball , we take t = 6.14 s

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