Problem 1 - Hunter and Monkey. A monkey hangs from a branch at a height h above

Question

Problem 1 - Hunter and Monkey. A monkey hangs from a branch at a height h above the
ground. A hunter stands on the ground at a distance R from the tree, and is pointing his gun at
the monkey at an angle (theta) to the horizontal, as shown in figure (1):

1. Let the speed of the bullet as it leaves the gun be v. Neglect air resistance. If th
monkey releases his grip at the same instant the hunter res the gun, show that the monkey will
be struck by the bullet after falling a distance (delta)y given by:

2.Show also that in order for this shot to work, the height of the tree and the distance
of the hunter to the tree must be related by:

No idea where to even start.. Please show all steps since I struggle greatly with physics!
Thank You!!!

Explanation / Answer

1.) First you need to find the time it takes the bullet to travel the horizontal distance R: horizontal velocity = v*cos(theta) distance = velocity * time R = v*cos(theta)*t t = R/(v*cos(theta)) Now, you need to find the distance the monkey falls in t seconds. delta y = Vi * t + (g / 2) t^2 delta y = 0 + (g / 2) * (R / (v*cos(theta)))^2 delta y = (g / 2) * (R^2 / (v^2 * cos(theta)^2)) here, we have to use some tricks to convert the cos(theta)^2 to something else we have... using sin^2 + cos^2 = 1, you can divide both sides by cos^2 to get: sin^2 / cos^2 + 1 = 1/cos^2, you know sin^2 / cos^2 = tan^2, so: tan^2 + 1 = 1 / cos^2 Now, putting that in for the 1/cos^2 we had above: delta y = (g / 2) * (R^2 * (1 + tan(theta)^2)) / (v^2) and since tan = opposite / adjacent, tan = h / R in our problem. So now we have: delta y = (g / 2) * ((R^2 + h^2) / v^2) 2.) This part I'm not sure where to start on, but hopefully part 1 helps a bit =)

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