Your monkey teammate needs to pass this challenge. Monkey has a mass of 80.0 kg.

Question

Your monkey teammate needs to pass this challenge. Monkey has a mass of 80.0 kg. She will be flying in a helicopter and jump at a height of delta-y1= 40 meters. The helicopter is flying at an angle of theta=25 degrees above horizontal at a speed of 18 m/s. The monkey must land on the ideal rocket skateboard resting on the horizontal plain. At the instant the monkey lands on the skateboard, she activates a thruster rocket that provides an acceleration of 2.25 m/s^2. The monkey and the skateboard become airborne at the cliff edge of the plain as the thruster shuts off. To survive this challenge, monkey and skateboard must land at the landing point and smoothly decelerate to a stop at the stopping point. The cliff is delta-y2= 25 meters above the landing point. Delta=x3= 85 m wide (bottomless chasm). Distance from landing point to stopping point is delta-x4= 160 m.

1)      Time monkey will be in the air before landing on skateboard?

2)      The horizontal distance, delta-x1, the skateboard must be from the point directly below the helicopter when the monkey jumps?

3)      The horizontal distance, delta-x2, the skateboard must be from the cliff if the monkey is to have a horizontal velocity sufficient as she leaves the cliff to clear the bottomless chasm and land at the landing point?

4)      The length of time that the acceleration of the skateboard must be applied so the monkey has the sufficient horizontal velocity to clear the chasm and land at landing point?

5)      Acceleration that must be applied to monkey and skateboard when a new thruster activates upon landing at the landing point, for them to stop at the stopping point?

6)      Time it will take for monkey and skateboard to stop?

Explanation / Answer

let the downward direction be +ve y direction

U = Ux i + Uy j

Ux = 18Cos25 = 16.3 m/s

Uy = -18Sin25 = -7.6 m/s

Then :

1) y1 = Uy*t + 0.5g*t^2

40 = (-7.6*t + 0.5*9.8*t^2

=> 4.9t^2 - 7.6t - 40 = 0

=> t1 = 3.73 sec

2) x1 = Ux*t1 = 16.3*3.73 = 60.85 m

3) Let horizontal velocity before leaving the cliff = V

and time for crossing bottomless chasm = t3

y3 = 0.5*g*(t3)^2

=> 25 = 0.5*9.8*t3^2

=> t3 = 2.26 sec

x3 = V*t3

=> V = x3/t3 = 85/2.26 = 37.63 m/s

So for attaining this final velocity V,

x2 = (V^2 - Ux^2)/(2*a)

here a = acceleration due to the thruster

x2 = (37.63^2 - 16.3^2) / 2*2.25 = 255.55m

4) t2 = (V - Ux) / a = (37.63 - 16.3)/2.25 = 9.48 sec

5)velocity at the point of landing will be V = 37.63 m/s in horizontal direction.

So, x4 = (0 - V^2)/2a'

=> a' = (-37.63^2)/(2*160) = -4.425 m/s^2

negative sign indicates acceleration in opposite direction

6) t4 = (0 - V)/(a) =-37.63/(-4.425) = 8.5 sec

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