A trained dolphin leaps from the water with an initial speed of 12 m/s. The trai

Question

A trained dolphin leaps from the water with an initial speed of 12 m/s. The trainer releases the ball at the horizontal distance of 6.42 m away and the vertical distance of 4.04 m above the water. Suppose the dolphin's launch angle is 45.0 ?

a.) What is the vertical distance between the dolphin and the ball when the dolphin reaches the horizontal position of the ball? We refer to this as the "miss distance."

b.) If the dolphin's launch speed is reduced, will the miss distance increase, decrease, or stay the same?

c.) Find the miss distance for a launch speed of 10.0 m/s.

Explanation / Answer

a)

consider the motion of dolphin

X = horizontal distance travelled by dolphin = 6.42 m

Vox = velocity of dolphin along the horizontal direction = 12 Cos45 = 8.5 m/s

t = time taken to travel the horizontal distance = ?

since along the horizontal direction , acceleration is zero ,

t = X/Vox

t = 6.42/8.5

t = 0.76 sec

Y = vertical position of the dolphin = ?

a = acceleration = - 9.8 m/s2

Voy = velocity of dolphin along the vertical direction = 12 Sin45 = 8.5 m/s

t = time of travel = 0.76 sec

using the equation

Y = Voy t + (0.5) a t2

Y = 8.5 (0.76) + (0.5) (- 9.8) (0.76)2

Y = 3.63 m

consider the motion of the ball along the vertical direction

Yob = initial position = 4.04 m

Yfb = final position = ?

Vob = initial velocity = 0 m/s

t = time = 0.76 sec

Using the equation

Yfb = Yob + Vob t + (0.5) a t2

Yfb = 4.04 + 0 (0.76) + (0.5) (- 9.8) (0.76)2

Yfb = 1.21 m

d = distance between dolphin and ball = Y - Yfb = 3.63 - 1.21 = 2.42 m

b)

the miss distance will decrease

this is because as the launch speed of the dophin is reduced , the value of Y will decrease . hence the value of miss distance "d" will decrease

c)

consider the motion of dolphin

X = horizontal distance travelled by dolphin = 6.42 m

Vox = velocity of dolphin along the horizontal direction = 10 Cos45 = 7.1 m/s

t = time taken to travel the horizontal distance = ?

since along the horizontal direction , acceleration is zero ,

t = X/Vox

t = 6.42/7.1

t = 0.90 sec

Y = vertical position of the dolphin = ?

a = acceleration = - 9.8 m/s2

Voy = velocity of dolphin along the vertical direction = 10 Sin45 = 7.1 m/s

t = time of travel = 0.90 sec

using the equation

Y = Voy t + (0.5) a t2

Y = 7.1 (0.90) + (0.5) (- 9.8) (0.90)2

Y = 2.42 m

consider the motion of the ball along the vertical direction

Yob = initial position = 4.04 m

Yfb = final position = ?

Vob = initial velocity = 0 m/s

t = time = 0.90 sec

Using the equation

Yfb = Yob + Vob t + (0.5) a t2

Yfb = 4.04 + 0 (0.90) + (0.5) (- 9.8) (0.90)2

Yfb = 0.071 m

d = distance between dolphin and ball = Y - Yfb = 2.42 - 0.071 = 2.35 m

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