A soccer ball is kicked with an initial horizontal velocity of 20 m/s and an initial vertical velocity of 16 m/s.

I've calculated:

Initial Speed: 25.61 m/s
Angle: 38.66 degrees
Maximum Height it reaches: 13.06 m
Distance kicked: 65.29m

NOW I NEED:

What is the speed of the ball 0.7 seconds after it was kicked?

How high above the ground is the ball 0.7 seconds after it is kicked?

help? please.

To find the speed of the ball 0.7 seconds after it was kicked, we can use the horizontal velocity component, as the vertical velocity does not affect the speed in this case.

Given:
Initial horizontal velocity (Vx) = 20 m/s

Since there is no horizontal acceleration, the horizontal speed remains constant. Therefore, the speed of the ball 0.7 seconds after it was kicked is also 20 m/s.

Next, let's calculate the height of the ball 0.7 seconds after it is kicked.

We know:
Initial vertical velocity (Vy) = 16 m/s
Time (t) = 0.7 seconds

To calculate the height, we can use the kinematic equation for vertical motion:

h = Vyt + (1/2)gt^2

where:
h is the height
Vy is the initial vertical velocity
t is the time
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the given values, we have:

h = (16 m/s)(0.7 s) + (1/2)(9.8 m/s^2)(0.7 s)^2

Calculating this expression:

h = 11.2 m + 2.015 m

Therefore, the height of the ball 0.7 seconds after it is kicked is approximately 13.215 m.

To find the speed of the ball 0.7 seconds after it was kicked, you can use the equations of motion. Since the initial velocity has both horizontal and vertical components, you need to consider both components separately.

Using the vertical component, you can determine the time of flight of the ball. The formula to calculate the time of flight is:

time of flight = (2 * initial vertical velocity) / gravitational acceleration

In this case, the initial vertical velocity is 16 m/s, and the gravitational acceleration is approximately 9.8 m/s^2. Substituting these values into the formula:

time of flight = (2 * 16) / 9.8 = 3.27 seconds

Now, since you want to calculate the speed and height of the ball at 0.7 seconds, you need to consider the horizontal and vertical motion separately.

For the horizontal motion, the speed remains constant throughout the motion. Since the ball was kicked with an initial horizontal velocity of 20 m/s, the speed of the ball at any time during its flight is also 20 m/s. Thus, the speed of the ball 0.7 seconds after it was kicked is 20 m/s.

For the vertical motion, you can calculate the height above the ground using the equation:

vertical displacement = (initial vertical velocity * time) + (0.5 * gravitational acceleration * time^2)

Substituting the values, in this case, the initial vertical velocity is 16 m/s, the time is 0.7 seconds, and the gravitational acceleration is approximately 9.8 m/s^2, you can calculate the height above the ground:

vertical displacement = (16 * 0.7) + (0.5 * 9.8 * (0.7)^2) = 11.92 meters

Therefore, the speed of the ball 0.7 seconds after it was kicked is 20 m/s, and the height above the ground is 11.92 meters.

the horizontal speed hs does not change

the vertical speed vs is 16-9.8t
vs(.7) = 16-9.8*.7 = 9.14

so, the speed is sqrt(9.14^2 + 20^2) = 21.99 m/s

vertical distance d = 1/2 at^2
d(.7) = 4.9*.7^2 = 2.4m