A golf ball is stuck at point o on the ground and moves with an initial velocity of 20ms-1 at an angle of 53 degrees to the horizontal.The ball subsequently lands at a point x which is on the same horizontal levelas o.

Question

a) show that the time taken by the ball to reach point x is approximately 3.26seconds.

b)calculate the distance OX.
I did cos(53)X20=12.04 and i think thats he answer for b

C)state
i)the least speed of the ball during its flight from o to x.
ii)the direction ofmotion of the ball when this least speed occurs

2) A golfer hits a ballfrom ground level on a horizontal surface. The initial velocity of the ball is 21ms-1 at an angle of 60 degrees above the horizontal. Assume that the ball is a particle and that no resistance forces act on the ball.

a)find the maximum height of the ball

b)find the range of the ball

c)find the speed of the ball at its maximum height

please help and give me simple detailled explanations as i have no clue

Answer 1

Vo = 20m/s[53o].
Xo = 20*Cos53 = 12.04 m/s.
Yo = 20*sin53 = 15.97 m/s.

a. Y = Yo + g*Tr = 0 @ max. ht.
15.97 - 9.8Tr = 0
Tr = 1.63 s. = Rise time.
Tf = Tr = 1.63 s. = Fall time.
Tr+Tf = 1.63 + 1.63 = 3.26 s. = Time to reach point X.

b. d = Xo*(Tr+Tf) = 12.04 * 3.26 = 39.25 m.

c. Vmin = Xo + Yi = 12.04 + 0i = 12.04m/s[0o].
Direction = 0o.

2. Vo = 21m/s[60o].
Xo = 21*Cos60 = 10.5 m/s.
Yo = 21*sin60 = 18.2 m/s.

a. Y^2 = Yo^2 + 2g*h.
0 = (18.2)^2 - 19.6h, h = ?.

b. Range = Vo^2*sin(2A)/g.
Vo = 21 m/s, A = 60o, g = 9.8 m/s^2.

c. Vmin = Xo + 0i = 10.5 m/s[0o].