Red Kangaroo leaped essentially vertically up from a cliff to achieve a maximum

Question

Red Kangaroo leaped essentially vertically up from a cliff to achieve a maximum height of 10 feet above the cliff and then fell below the cliff. There was no wind or air resistance. A) What was the initial vertical velocity of the Red Kangaroo? B) What was the velocity of the Kangaroo 5 feet above the cliff? C) At what heights was the instantaneous velocity of the Kangaroo one-half of the initial vertical velocity? D) At what vertical position was the kangaroo when the absolute value of the vertical velocity was i) 0 m/s and ii) the initial vertical velocity? E) What was the instantaneous velocity of the kangaroo when the kangaroo was i) 5 feet below the cliff and ii) 10 feet below the cliff F) How far below the cliff was the kangaroo if the absolute value of his final instantaneous velocity was twice the initial instantaneous velocity?

Explanation / Answer

For easiness, convert feet into meters.

10 feet = 3.05 meter

5 feet = 1.52 meter.

(A) Suppose the initial velocity of the kangaroo is u m/s.

apply the expression -

v^2 = u^2 + 2*a*s

=> 0 = u^2 - 2*9.81*3.05

=> u^2 = 59.841

=> u = 7.74 m/s

(B) Consider the velocity of the kangaroo 5 ft above the cliff is v m/s.

Now put the values in the above expression -

v^2 = 7.74^2 - 2*9.81*1.52

=> v^2 =59.84 - 29.82 = 30.02

=> v = 5.48 m/s

(C) (7.74/2)^2 = 7.74^2 - 2*9.81*h

=> 19.62*h = 44.9307

=> h = 2.29 m

(D) (i) v^2 = 7.74^2 + 2*9.81*1.52 = 89.73

=> v = 9.47 m/s.

(ii) like-wise -

v^2 = 7.74^2 + 2*9.81*3.05 = 119.7486

=> v = 10.94 m/s.

(F) In that case -

(2*7.74)^2 = 7.74^2 + 2*9.81*h

=> 19.62*h = 179.7228

=> h = 9.16 m.

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