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 6.49 m away and the vertical distance of 4.06 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." delta h=?m

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

Answer: stay the same.

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

Answer 1

for dolphin

PROJECTILE

along horizontal
________________

initial velocity v0x = v*costheta

acceleration ax = 0

initial position = xo = 0

final position = x

displacement = x - x0

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - x0 = v*costheta*T

T = (x - x0)/(v*costheta)

T = x/(v*costheta)

T = 0.765 s

along vertical
______________

initial velocity v0y = v*sintheta

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 0

final position y = 0

from equation of motion

y-y0 = v0y*T + 0.5*ay*T^2

for the ball

y - h = (1/2)*g*T^2

y - h = (1/2)*g*T^2

y = h + (1/2)*g*T^2

miss distance = v0y*T + 0.5*ay*T^2 - h - (1/2)*g*T^2

miss distance = voy*T - h = v*sintheta*x/(v*costheta) - h

miss distance = x*tantheta - h

miss distance = (6.49*tan45) - 4.06 = 2.43 m

(b)

stay the same

dose not depend on the speed v

(c)

miss distance = (6.49*tan45) - 4.06 = 2.43 m