# Monkey and Hunter:Projectile Motion

1. Oct 11, 2014

### TheDestroyer123

1. The problem statement, all variables and given/known data

Consider the setup of a gun aimed at a target as shown in the figure. The target is to be dropped from point A at t = 0 s, the same moment as the gun is fired. The bullet hits the target at a point P which is along vertical line to the ground. Let the initial speed of the bullet be Vi= 149 m/s and the angle between the vector Vi and the x-axis be θ = 61.4 degrees and the starting height of the target be 87.3 m. The horizontal distance lies on the x-axis.
The acceleration of gravity is 9.8 m/s2.

a) Determine the y component of the velocity of the bullet at the time of collision.

b) At what height, above the ground, do the bullet and target hit.

c) Find the speed of the target at the point of collision.

2. Relevant equations

-None-

3. The attempt at a solution

I believe I am correct for these answers, but when I enter them into a website where we submit our work, it says the answers are wrong for everything (a-c). Am I doing something wrong?

a) Vyimpact=Vyi+at
Delta x = 87.3tan(61.4)
From that: 160.1195=149cos(61.4)t
t=2.2449

Vyimpact=Vyi+at
Vyimpact= 149sin(61.4)-(9.8)(2.2449)=108.819 m/s

b) Distance target fell in 2.2449 seconds= -(1/2)(a)t2
delta y= -(.5)(9.8)(2.2449)2 = -24.6939

Distance target was hit above ground = 87.3-24.6939=62.6061 m

c) Vxi=Vx2 *Neglecting any horizontal factors
Vx2=149cos(61.4)=71.3251 m/s
Vy2=Vyimpact
Vy2= 108.819

Vimpact= sqrt(71.32512 + 108.8192)= 130.1108 m/s

2. Oct 11, 2014

### TSny

Hello and welcome to PF!

Check to see if you used the trig function correctly here.
Otherwise, your work looks good to me. :)