Question

  A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 19.6 m/s at an angle of 50.0? 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

horizontal component of velocity :

velocity vx = vcos?

= (19.6 m/s)cos50

= 12.5986 m/s

vertical component of the velocity :

velocity vy = vsin?

= (19.6 m/s)sin50

= 15.0145 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

= (15.0145 m/s)2/2(9.8 m/s2)

= 11.5018 m

therefore , ?y = 11.5018 m - 2.8 m

= 8.7018 m

from kinematic equations,

vfy2 = v0y2 + 2as

vfy2 = (0 m/s)2 + 2g(?y)

vfy2 = 2(9.8 m/s2)(8.7018 m)

vfy = 13.0597 m/s

therefore , resultant velocity is

v = ?[(12.5986 m/s)2 + (13.0597 m/s)2]

= 18.146 m/s

= 18.15 m/s.....................ans

Answer 2

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else).

A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 17.7 m/s at an angle of 41.0? 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.

initial velocity of the ball

Ux = 17.7*cos41= 13.36 m/s

Uy= 17.7*sin41 =11.61m/s

v changes along y direction, hence Vx = Ux

s=Uy*t -0.5gt²

s= 2.8 ,

on solving we get t=2.097 s and t= 0.27s

as it is falling down we should take the larger t

hence t=2.097

Vy = Uy - gt

=-8.88 m/s

speed = magnitude of V = ?(Vx² + Vy²)

= 16.04 m/s   

Answer 3

K = K + P
(1/2)mv0^2 = (1/2)mv^2 + mgh
v0^2 = v^2 + 2gh
v^2 = v0^2 - 2gh
v = sqrt(v0^2 - 2gh)
v = sqrt(19.6^2 - 2(9.8)(3.8))
v = 17.59 m/s