Acannon sends a projectile towards a target a distance 1420 m away. the initial velocity makes an angle 35◦ with the horizontal. the target is hit. the acceleration of gravity is 9.8 m/s2 . what is the magnitude of the initial veloc- ity?

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

From the equation of x-position we know that

Solving for

(1)

Now, if we analyze the equation of y-position we got

At the end of the motion y=0

Knowing the equation for in (1)

Solving for t

Now, we can solve (1)

121.7 m/s

Explanation:

Horizontal range, R = 1420 m

Angle of projection, θ = 35°

acceleration due to gravity, g = 9.8 m/s^2

let u be the velocity of projection.

Use the formula for horizontal range

u = 121.7 m/s

Thus, the velocity of projection is 121.7 m/s.

2NH3 --> N2 + 3H2 decomposition