Aball is kicked from ground level with a velocity of 20 m/s at 30 degrees above the horizontal. determine the horizontal displacement to where it lands and find how high the ball went at max height.

the horizontal displacement is 34.64 meters

The maximum height is 15 meters

Step-by-step explanation:

As the ball moves from the ground at 30 degrees, it travels through a parabolic path. It can therefore be likened to a projectile. The horizontal displacement to where it lands is the range. The formula for determining range,R is

R = U^2Sin2#)/g

Where

U represents the initial velocity of the ball

g is acceleration due to gravity

# represents the angle above the horizontal.

From the information given

# = 30 degrees

U = 20 m/s

g = 10m/s^2

R = 20^2Sin2×30)/10 = (400Sin60)/20 = (400×0.8660)/10

= 34.64 meters

To determine the maximum height of the ball, H,

H = (U^2Sin^2#)/2g

H = (20^2 × Sin^2 30)/2×10

H = (400 × 0.866^2)/20 = 15 meters

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.