The golfer must hit the ball through an opening of a door that as a good distanc

Question

The golfer must hit the ball through an opening of a door that as a good distance away. The Catch is that a door immediately starts to close over the opening as soon as the ball is struck. Assuming the golfer can hit the ball with an initial velocity of 40 m/s. The height of the opening is initially 25 m and the golfer is a distance of 100 m from the wall. The wall begins to move with a velocity of 6 m/s the instant the ball is struck.What launch angles %u03B8(theta) will allow the ball to pass through the part of the opening that isn't yet covered.

a.) Calculate how long it takes the ball to reach the wall.

b.) Calculate the height of the ball when it reaches the wall.

Initiall Velocity of the ball = 40 m/s

Velocity of the wall = 6 m/s

L1=14 m

L2=25 m

L3=2 m

L4=100 m

Explanation / Answer

Let the horizontal and vertical componet of velocity of ball be vx and vy just after hitting it

Let the ball just crosses the wall at time t

heigth of ball = L1 - L3 + 6t = 14-2 + 6t = 12 + 6t  

now height = vy * t - 0.5 * 9.8 * t^2

12 + 6t = vy*t - 4.9 t^2

vy = 12/t + 6 + 4.9t

horizontal distance covered by ball = L4 = 100 m = vx * t

vx = 100 / t

now given ball has an initial velocity of 40 m/sec

so, vx^2 + vy^2 = 40^2

10000/t^2 + (12 + 6t + 4.9t^2 )^2 / t^2 = 1600

so,

10000 + ( 12 + 6t + 4.9t^2)^2 = 1600 t^2

10000 + 144 + 36t^2 + 24.01 t^4 + 144t + 117.6 t^2 + 58.8 t^3 = 1600t^2

24.01 t^4 + 58.8 t^3 -1446.4t^2   + 144 t + 10144

t =5.670837251156119 secs   and t = 3.25772226840156

( ignore the -ve values as time cannot be negativr )

( use the following calculator to solve the above equation : http://easycalculation.com/algebra/quartic-equation.php )

So there are two cases possible

a)

case I -- t = 5.670837251156119 secs

vx = 100 / t=   100 /5.670837251156119= 17.6340804 m/sec

vy = (12 + 6t + 4.9t^2 ) / t   = 35.90319218 m/sec

tan theeta = vy / vx

so theeta = 63.8417 degrees

case II - t = 3.25772226840156 secs

vx = 100 / t=   100 /3.25772226840156 = 30.6963 m/sec

vy = (12 + 6t + 4.9t^2 ) / t   = 25.6464 m/sec

tan theeta = vy / vx

so theeta =39.878356 degrees

b) heigth of ball when it reached wall = 12 + 6 *t

case I   --- t = 5.670837251156119 secs

heigth = 12 + 6 * 5.670837251156119 = 46.0250235069367 metress from ground

case II - t = 3.25772226840156 secs

heitght = 12 + 6*t = 31.54633361040936 m from the ground

but L1 + L2 = 12 + 25   = 39 metres ..

so the heigth of the ball can be at a maximum heigth of 39 metres from ground..

so ,.. case I is not possible ..

so only case II is possible ..

a) time = 3.25772226840156 secs

angle = 39.878356 degrees

b) heigth = 31.54633361040936 m

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