projectile motion question?

Question

A projectile is fired horizontally from a gun that is 41.0 m above flat ground, emerging from the gun with a speed of 250 m/s.
It remains in the air for 2.9 seconds.

1.) At what horizontal distance from the firing point does it strike the ground? (m)

2.) What is the magnitude of the vertical component of its velocity as it strikes the ground? (m/s)

Answer 1

The key point to remember is that the vertical and horizontal components can be calculated completely independant of one another. For the x component:

X=Xo+(Vox)t

Xo is the initial x position and is zero. Vox is the x component of the initial velocity and is equal to Vo*cos(theta-naught) where Vo is the magnitude of the velocity and theta-naught is the initial angle. They give you the time of flight and the initial angle, so:

X=250*cos(0)*2.9

X=725 m

For the y, we must use the following:

V(y)=Voy-gt

Voy is the y component of the initial velocity and is equal to Vo*sin(theta-naught)

V(y)=250sin(90)-9.8*2.9

V(y)=0-28.42

=-28.42 m/s

Answer 2

horizontal distance = vo*t = 250*2.9 = 725m
vertical distance = 1/2gt^2 counting from the position of the gun
which is coherent with the given time
Observe that giving the time is NOT needed as
1/2gt^2=41 gives you t = sqrt( 82/g) =2.9 s
v= gt (module) = 28.42 m/s vertical component directed downwards

Answer 3

Given:
dx = 41.0m
vi = 250m
t = 2.9 s

Find:
dx = ?
vfy = ?

Solution:
1)
d = vt
d = 250 m(2.9s)
d = 725 m

2)
vfy = gt
vfy = (9.8m/s^2)(2.9s)
vfy = 28.42 m/s

Good luck and God bless!