Answer: The initial velocity of the projectile is 3.1 km/s
Explanation:
There is a problem with the question because we do not know in which direction the projectile travels 70 km.
I will assume that the projectile travels 70 km horizontally, as is the usual case.
I will also assume that 25 seconds after the projectile is fired, it hits the ground.
Suppose that the initial velocity of the projectile is V0, and this velocity V0 can be thought as the hypotenuse of a triangle rectangle, then the x-component and the y-component of the velocity will be the catheti of this triangle rectangle, then we can write:
Vy = V0*sin(25°)
Vx = V0*cos(25°).
Now, the vertical problem does not affect the horizontal one, so we can see only the horizontal case.
Here we don't have any force, then there is no acceleration, then the velocity is constant and it is equal to:
v(t) = V0*cos(25°)
For the horizontal position we can integrate over time to get:
p(t) = V0*cos(25°)*t + p0
where p0 is the initial position, and we can assume that it is equal to zero.
Then:
p(t) = V0*cos(25°)*t
Now we know that in t = 25s, the position is 70km
(because the projectile travelled 70km in 25 seconds) then:
p(25s) = 70km = V0*cos(25°)*25s
Now we can solve this for V0.
V0 = 70km/(cos(25°)*25s) = 3.1 km/s
The initial velocity of the projectile is 3.1 km/s
The thing you can get from this is that we can analyze the vertical problem and the horizontal problem separately (for example, here we only solved the horizontal case and we ignored the vertical one).
