Aprojectile is fired from the origin (at y = 0 m) as shown in the diagram. the initial velocity components are v0x=310m/s and v0y=26m/s the projectile reaches maximum height at point p, then it falls and strikes the ground at point q, which is 20 m below the launch point. what is the horizontal distance that the projectile travels (labeled x in the diagram)?

700 m

3.2 km

870 m

1.3 km

1.9 km

1.9 km

Explanation:

The projectile would take the same time to cover distance x as it would take to reach the ground.

Let us take upward direction as positive and downward direction as negative.

The net displacement in vertical direction is Δy = -20 m. The projectile falls down under gravity.

we will find out the time taken for projectile to reach the ground.

Using second equation of motion:

Δy = Uoy t + 0.5 a t²

-20 m = 26 m/s × t + 0.5 × -9.8 m/s² × t²

⇒ 4.9 t²-26 t-20 = 0

Solving the quadratic equation and neglecting the negative value,

t = 5.98 s.

In the same time, horizontal distance covered would be:

x = Vox t = 310 m/s × 5.98 s = 1853.8 m ≈ 1.9 km

massxgravity x heght=weightxheight= pe

4900=490xh

h=10