A golfer, standing on a fairway, hits a shot to a green that is elevated 5.30 m

Question

A golfer, standing on a fairway, hits a shot to a green that is elevated 5.30 m above the point where she is standing. The ball leaves her club at an angle of 36.0° above the ground, with a speed of 31.0 m/s. Find the time that the ball is in the air before it hits the green. How far did the ball travel horizontally? Determine the ball's speed right before it hits the green. A golfer, standing on a fairway, hits a shot to a green that is elevated 5.30 m above the point where she is standing. The ball leaves her club at an angle of 36.0° above the ground, with a speed of 31.0 m/s. Find the time that the ball is in the air before it hits the green. How far did the ball travel horizontally? Determine the ball's speed right before it hits the green. A golfer, standing on a fairway, hits a shot to a green that is elevated 5.30 m above the point where she is standing. The ball leaves her club at an angle of 36.0° above the ground, with a speed of 31.0 m/s. Find the time that the ball is in the air before it hits the green. How far did the ball travel horizontally? Determine the ball's speed right before it hits the green.

Explanation / Answer

here,

initial speed , u = 31 m/s

theta = 36 degree

height , h = 5.3 m

let the horizontal distance be x

using equation of trajectory

y = x * tan(theta) - g * x^2 /( 2 * ( u * cos(theta))^2)

5.3 = x * tan(36) - 9.81 * x^2 /(2 * ( 31 * cos(36))^2)

solving for x

x = 8 m

let the time taken be t

x = u * cos(theta) * t

8 = 31 * cos(36) * t

t = 0.32 s

the final vertical speed , vy = u * sin(theta) - g * t

vy = 31 * sin(36) - 9.81 * 0.32

vy = 15.1 m/s

the ball's final speed , v = sqrt(vy^2 + (u * cos(theta))^2)

v = 29.3 m/s

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