Question

Let's say that the dart is launched with initial velocity components of 28.0 m/s horizontally and 6.72 m/s vertically. Imagine that this is taking place on a planet where the acceleration due to gravity is g = 9.00 m/s, and neglect air resistance.

What is the distance between the dart and the monkey at the instant the dart passes directly below the monkey? Note that the simulation shows that the monkey starts 40 m higher, and 100 m horizontally, from where the dart is launched, and you can make use of those numbers in your calculation.
_______ m

Answer 1

Let's solve the problem by relative motion method.

Since the monkey and dart have the same acceleration due to gravity,

Hence, acceleration of dart with respect to monkey will be,

(Acceleration of dart with respect to ground) - ( acceleration of monkey with respect to ground) =0

Hence the motion with respect to monkey is a non accelerated motion.

Now, initial velocity of the dart with respect to monkey will be same as that with respect to ground, since the initial velocity of monkey is zero.

Hence horizontal velocity of dart with respect to monkey is 28m/s

And vertical velocity of dart with respect to monkey is 6.72m/s

Now, time taken by the dart to reach vertically below the monkey will be given by horizontal distance covered divided by the speed in horizontal direction.

i.e. (100m)/(28m/s) seconds

And in the calculated time the dart moves vertically up by a distance of

(6.72m/s)*(100m)/(28m/s) = 24m with respect to monkey.

But initially the monkey was 40 m above the dart, hence after travelling 24m vertically the dart is still 16m below the monkey.

Hence answer is dart will be 16m below the monkey.