A soccer ball is kicked from the ground with initial speed 17 m/s at angle 20 0

find
a) The horizontal distance that the ball travels after 4 s ?
b) The speed of the ball at the highest point ?

at the highest point, it has only horizonal speed, 17cosine20

distance:
time in air is found by
hf=hi+vi'(t)-4.8t^2
or t=vi'/4.8=17sin20/4.8

distance=17cos20*17*sin20/4.8

To solve these problems, let's first break down the initial velocity into its horizontal and vertical components.

Given:
Initial speed = 17 m/s
Launch angle = 20 degrees

Converting the angle from degrees to radians:
Launch angle in radians = 20 * (π/180) ≈ 0.349 radians

Using the trigonometric relationships, we can find the horizontal and vertical components of the initial velocity:

Horizontal component:
Vx = V * cos(θ)
Vx = 17 * cos(0.349)
Vx ≈ 17 * 0.9397
Vx ≈ 15.98 m/s

Vertical component:
Vy = V * sin(θ)
Vy = 17 * sin(0.349)
Vy ≈ 17 * 0.3420
Vy ≈ 5.81 m/s

a) Finding the horizontal distance traveled after 4 seconds:
We can use the horizontal component of velocity to calculate the distance since horizontal velocity is constant.

Horizontal distance (d) = Vx * time
d = 15.98 * 4
d ≈ 63.92 meters

Therefore, the horizontal distance traveled by the ball after 4 seconds is approximately 63.92 meters.

b) Finding the speed of the ball at the highest point:
At the highest point of the projectile's trajectory, the vertical velocity reaches zero. The speed at this point is equal to the horizontal component of velocity.

Speed at highest point = Vx
Speed at highest point ≈ 15.98 m/s

Therefore, the speed of the ball at the highest point is approximately 15.98 m/s.

To find the horizontal distance that the ball travels after 4 seconds, you can use the horizontal component of the initial velocity.

a) The horizontal component of the initial velocity can be found using the following formula:

Vx = V * cos(theta)

Where Vx is the horizontal component of velocity, V is the initial speed (17 m/s), and theta is the launch angle (20 degrees).

Plugging in the given values, we get:

Vx = 17 m/s * cos(20 degrees)

Vx = 17 m/s * 0.9397

Vx = 15.9539 m/s

This is the horizontal speed, or the speed in the x-direction.

To find the horizontal distance, we can use the formula:

Distance = Velocity * Time

Plugging in the values, we have:

Distance = 15.9539 m/s * 4 s

Distance = 63.8156 m

Therefore, the horizontal distance that the ball travels after 4 seconds is approximately 63.82 meters.

b) To find the speed of the ball at the highest point, we need to find the vertical component of the velocity at that point.

The vertical component of the initial velocity can be found using the following formula:

Vy = V * sin(theta)

Where Vy is the vertical component of velocity, V is the initial speed (17 m/s), and theta is the launch angle (20 degrees).

Plugging in the given values, we get:

Vy = 17 m/s * sin(20 degrees)

Vy = 17 m/s * 0.3420

Vy = 5.814 m/s

Since the velocity at the highest point is purely vertical, the speed at the highest point is equal to Vy. Therefore, the speed of the ball at the highest point is approximately 5.81 m/s.