Here is a physics question dealing w/ two dimensional motion?

Question

A golfer wishes to chip a shot into a hole 40 m away on flat level ground. If the ball sails off at 40°, what speed must it have initially? Ignore aerodynamic effects.

Answer 1

I assume you want the ball velocity in the direction of motion. Call that Vb. The horizontal component of that velocity is Vx, where

Vx = Vb cos(40) .....(1)

The vertical component is

Vy = Vb sin(40) .....(2)

The ball must travel 40m in the time it stays aloft, Ta; so its horizontal velocity must be

Vx = 40/Ta .....(3)

The time it stays aloft is twice the time it takes to reach the peak of its trajectory, Tp. The peak is reached when

(Vy-gTp)=0; Tp = Vy/g, so Ta = 2*Vy/g .....(4)

Now plugging Ta from Eq (4) into Eq (3) you get

Vx = (40g)/(2Vy) .....(5)

Now plugging Eq (1) for Vx and Eq (2) for Vy into Eq (5), we get


Vbcos(40) = (40g)/(2Vbsin40) .....(6)

Solve Vb to get

Vb = sqrt[(40g)/(2sin40cos40)] .....(7)

This gives Vb = 20 m/sec or 72 km/hr

Answer 2

The initial velocity of the ball is the vector sum of its horizontal and vertical velocity components:

vxi = vi * cos (40)
vyi = vi * sin (40)

(vyi / vxi) = tan(40)

Since it's two-dimensional motion, you have to analyze it in two equations to find the vxi and vyi:

Horizonatal (or x-axis):
There is no accleeration in the x-driection, so, ax = 0 and vxi is constant.
x(t) = vx*t + x0 = (vxi) * t + 0 = 40

Vertical (or y-axis):
Gravity is causing acceleration in the y-direction, so ay = -9.8m/sec^2.
y(t) = (1/2)(-9.8)(t^2) + (vyi) * t = 0 (when the ball lands)

But, t = (40 / vxi) from above

(1/2)(-9.8)(1600/ (vxi)^2) + (vyi) * (40 / vxi) = 0
7840 / (vxi)^2 = (40)(vyi / vxi)
196 = (vxi) * (vyi)

But, from above, (vyi / vxi) = tan(40)
so, 196 = (vxi)*(vxi)*(tan(40))
vxi = 15.3 m/s
vyi = 12.8 m/s

vi = 20 m/s

Answer 3

I'd use an 8 iron. Choke up on it.

Answer 4

around 128.6 KMH

Answer 5

jp4rts is absolutely right.

good job

thanx

Answer 6

about 128.6 kmh