Summer homework physics problem?

Question

This is dumb, but I can't figure it out. Please help!

Tiger Woods, practicing golf on vacation, hits a golf ball off of the top of his hotel (35 m above ground level) with a velocity of 130 miles per hour. The golf ball is initially moving upward at a 42 degree angle (from horizontal).

a. Calculate the location where the golf ball lands.

b. Calculate the time of flight.

c. Calculate the maximum height reached by the ball.

d. Calculate the height of the ball two-thirds of the way to the landing spot (2/3 of the horizontal displacement, not 2/3 of the time of flight).

Sorry for the ridiculous question, but this is part of my summer homework and I can't even figure out how to set up this problem. If anyone can help me with any of the parts, thank you SO much!!

Answer 1

Kirah--

Please check my "numbers", but this is the correct method to do this problem.

This is a "typical" "Big 4 equations" motion problem.  The equations that we will use are...

In the horizontal direction (where we neglect air resistance and any other forces): distance = rate * time

In the vertical direction:

distance = 1/2 a t^2 +vo*t + Initial distance

vf^2 = vo^2 + 2 a distance

distance = 1/2 (vo+vf) * t

vf = vo + a*t

we need any three of {distance, v0, vf, a, t} to find them all.  

We must resolve the initial velocity vector into horizontal and vertical components...

Let's do stuff in meters, OK?

130 mph = 58.1 m/s

vo (vertical) = 58.1 * sin (42) = 38.9 m/s and

vo (horizontal) = 58.1 * cos (42) = 43.2 m/s

Now let's solve the motion of the golf ball from launch to its highest point:

initial distance = 35m (not distance, mind you)

vo = 38.9 m/s

a= -9.8 m/s^2 (negative because it is DOWN)

vf = 0 (it stops for a split second at the top before it comes down)

time = 3.97 sec and max height = 77.2 + 35 = 112.2 meters

Now from this point down to the landing...

vo=0

distance = 112.2 m

a=9.8 m/s^2 (I am defining down as positive this time).

Solving our vertical equations, gives...

vf = 46.9 m/s and time = 4.79 sec

The total time of flight = 4.79 (part 2) + 3.97 (part 1) = 8.76 seconds

Now we can get the range using d=v0 (horizontal) * time (total)

Range = 43.2 m/s * 8.76 sec = 378 meters (A monster drive)

The landing velocity is a vector whose magnitude we get from the Pythagorean thm:  v (landing) = SQRT(43.2^2 + 46.9 ^2) = 63.8 m/s and at an angle from the vertical of...arctan(43.2/46.9) = 42.6 degrees...PLEASE draw a picture to "see" this clearly.

======================================...

We can write the parabolic trajectory using parametric equation in terms of time:

y= -1/2 g t^2 + v0 t + 35 = -4.9 t^2 + 38.9 t + 35

and x = 43.2 t

we want to know what is going on at the 2/3 x point, i.e., x=2/3 * 378 = 252 = 43.2 * t thus time = 5.83 seconds.

Now we can easily find the height, by plugging in this value of t into the y equation to get...

y = height at 2/3 range 95.2 meters.

======================================...

That's all folks!!

Try to study this method and understand WHAT I did and why.  You will quickly master this stuff with a little practice.  Good luck.

-Fred

Recap: I "broke up" the velocity into x and y parts.

I used the vertical (y) equations twice to get the total time and the max height

I used the old familiar horizontal equation to get range (there is a "closed form" formula for range, but it is for level trajectories not those that start on a hill or valley).

I used vector trig. to get the landing velocity (speed and an angle).  Lastly...

I then introduced the parametric equation of the trajectory which allows "me" to find the position of the ball given either x or the time.

OK?  You will do great!!

Answer 2

First, convert 130 mph (Vo) to m/s

a)  y = [(-g*sec²Θ)/(2Vo²)]*x² + x*tanΘ + Ho  where y = 0 and Ho = 35m.  Solve for x, the horizontal distance from the tee; the + value will be your answer.

b)  y = Vo*t*sinΘ -½g*t² where y = -35.  Solve for t.

c)  h = (Vo*sinΘ)²/(2g)   (above tee)

d)  Use the eq in a) and put in an x value 2/3 of the one you calculated there.  The resulting height (y) will be relative to the tee (probably a - value)