Chasing the banana?

Classical Mechanics Level 3

While playing with his monkey, Tom threw a banana (horizontally) from a $45 \text{ m}$ high building while the monkey was running away from the building.

At the moment when banana was thrown speed of monkey was $4 \text{ m/s}$ and acceleration was $2 \text{ m/s}^2$ . Acceleration was constant throughout the motion and when banana was thrown monkey was at distance of $45\sqrt{2} \text{ m}$ away from the Tom.

Find the speed of projection if monkey catches banana just before it touches the ground.

Details and Assumptions :
1) Acceleration due to gravity is $g = 10 \text{ m/s}^2$ .
2) Banana and monkey are particles.
3) Write your answer of speed in $\text{ m/s}$ .

The answer is 22.

1 solution

Charlz Charlizard
Jul 1, 2015

45=(1/2) g t^2.. So t=3 sec..Let's say Projection speed is X..So (3) (X)=horizontal distance covered by the banana..Now V-4=(2) (3) :.V=10 m/s..Now horizontal distance covered by monkey=[(10) (10)-(4) (4)]/4=21 m..Now monkey is (45) [sqrt(2)] distance from Tom..implies it is 45 m distance away from the building :.45+21=3 X So X=22 m/s..

Chasing the banana?

The answer is 22.

1 solution

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