Question

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 18.1 m/s at an angle of 49.0˚ 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.

Answer 1

Solution,

Given,

Velocity, v = 18.1 m/s

Angle, theta = 49 degree

Horizontal component of velocity, vx = vcosθ

= 18.1 cos49 = 11.87 m/s

Vertical component of the velocity, vy = vsinθ

= 18.1 sin49 = 13.66 m/s

At maximum height, vy = 0 m/s

Using 3rd equation of motion,

v^2 = v0y^2 + 2as

0 = v0y^2 + 2ay

0 = v0y^2 + 2(-g)y

y = (v0y)^2/2g = 13.66^2/(2 x 9.8) = 9.52 m

therefore , Δy = 9.52 m - 3.2 m = 6.32 m

From 3rd equation of motion

vfy^2 = v0y^2 + 2as

vfy^2 = 0^2 + 2g(Δy)

vfy^2 = 2(9.8)(6.32)

vfy = 11.13 m/s

So, resultant velocity, v = sqrt [(11.87)^2 + (11.13)^2]   

V = 16.27 m/s

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