Find horizontal distance with given incline angle and acceleration...?

Question

I'm having some problems with this problem. Could someone please assist me? Thank you so much!

A dart gun is fired while being held horizontally at a height of .816 m above ground level, and at rest relative to the ground. The dart from the gun travels a horizontal distance of 7.65 m. A child holds the same gun in a horizontal position while sliding down a 45.8 degree incline at a constant speed of 1.94 m/s. The acceleration of gravity is 9.8 m/s^2.

What horizontal distance x will the dart travel if the child fires the gun forward when it is .539 m above the ground? Answer in units of m.

Answer 1

First step is to determine the muzzle velocity of the gun.

In the horizontal direction
7.65=v*t
and
.816=.5*9.8*t^2

solve for t
t=sqrt(2*.816/9.8)

t=.41 seconds

v=7.65/4.1

v=18.7 m/s

Now the dynamics of the child firing the gun
I will assume the child fires the gun horizontally
the child has horizontal velocity of 1.94*cos(45.8)
and vertical velocity of -1.94*sin(45.8)
the gun will have horizontal velocity of
18.7+1.94*cos(45.8)
vx=20.05 m/s
vy=-1.39-9.8*t

y(t)=.539-1.39*t-.5*9.8*t^2

when y(t)=0, the dart strikes the ground

t=.219 seconds to impact.
The horizontal distance then is
20.05*.219
4.39 m

j

Answer 2

A. Mechanical paintings is rigidity*distance W=118*a million.5=177J B. The paintings of the gravity is the exchange in ability power W=-m*g*h=19*9.80 one*a million.5*sin(25)=-118.1J (circulate is opposite to the path of the gravitation!) C. of course 0. The vector is perpendicular to the path of circulate. D. power conserves. the internet paintings of all forces would be enter paintings of the worker + paintings of gravitation. W=177+(-118.a million)=fifty 8.9J that quantity of power is obviously the exchange in kinetic power (crate is speeded up because it strikes).