Question

A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.5 m/s, and the distance from the limb to the level of the saddle is 3.26 m. (a) What must be the horizontal distance between the saddle and lmb when the ranch hand makes his move? 13.44 Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) For what time interval is he in the air? 1.28 Your response differs from the correct answer by more than 10%. Double check your calculations Noed Help? Resen

Answer 1

first we will solve part B

Using 2nd kinematic equation:

h = U*t + 0.5*a*t^2

h = distance between limb and saddle = -3.26 m

U = initial velocity of ranch hand = 0 m/sec, since dropped

a = -g = -9.81 m/sec^2

So,

-3.26 = 0*t + 0.5*(-9.81)*t^2

t = sqrt (3.26/(0.5*9.81))

t = 0.815 sec (Ans of part B)

Part A.

Now horizontal distance traveled by horse in this time will be

R = horizontal distance = Velocity of horse*time

R = (10.5 m/sec)*(0.815 sec)

R = 8.56 m = horizontal distance between saddle and limb (Ans of part A)

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