Monkey and Hunter:Projectile Motion

[TheDestroyer123]

Homework Statement

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 V_i= 149 m/s and the angle between the vector V_i 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.
Answer in units of m/s

b) At what height, above the ground, do the bullet and target hit.
Answer in units of m

c) Find the speed of the target at the point of collision.
Answer in units of m/s

Homework Equations

-None-

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) V_yimpact=V_yi+at
Delta x = 87.3tan(61.4)
From that: 160.1195=149cos(61.4)t
t=2.2449

V_yimpact=V_yi+at
V_yimpact= 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) V_xi=V_x2 *Neglecting any horizontal factors
V_x2=149cos(61.4)=71.3251 m/s
V_y2=V_yimpact
V_y2= 108.819

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

 

[TSny]

Hello and welcome to PF!

Delta x = 87.3tan(61.4)

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