a ball kicked with a 40 degree angle above the ground with velocity of 20 m/s, how high will it go

Question

Before claculating, break the original velocity into components. Calculate the VErtical component

Answer 1

The angle ( θ ) = 40 degree

Velocity = 20 m / s .

We first need to break the original velocity into components.

sinθ = opp/hyp

opp = sinθ * (hyp)                       

= sin 40° (20m/s)

= 0 . 6427 * (20m/s)

= 12 . 8557

opp = v_y = 12 .85 m/s

Calculate the vertical component :

At its maximum height, halways through its flight, the object won't be going up or down, so we'll

say that its final velocity at that point is zero.

v_f = 0 , v_i = 12 .85 m/s and  a = 9.81 m / s

Therefore v_f² = v_i² + 2ad

v_f² - v_i² = 2ad

d = (v_f² - v_i²) / 2a

Substitute the v_f ,v_i and a values in d = (v_f² - v_i²) / 2a

= [ 0² - (12 .85 )²] / 2(- 9.81)

=  - (12 .85 )² / - ( 19 .62)

= ( 165.1225)  / (19.62)

= 8 .416.

d = 8.4 m

The ball will reach a maximum height of 8.4 m.