Question

A soccer ball is kicked at an angle of 40° from the ground at a speed of 30 m/s. 1. Describe which motion (horizontal of vertical) of the ball is constant and which is accelerated

2. What are the vertical and horizontal components of the ball's initial velocity? 3. For what length of time is the ball in the air assuming the ground is flat and horizontal?
4. What is the range of the ball assuming the ground is flat and horizontal?​

Answer 1

(1). This motion is projectile motion.

(2). The horizontal initial velocity is 19.3 m/s.

The vertical initial velocity is 22.9 m/s.

(3). The time is 3.9 sec.

(4). The range of the ball assuming the ground is flat and horizontal is 90.4 m.

Step-by-step explanation:

Given that,

Angle = 40°

Velocity = 30 m/s

(1). A soccer ball is kicked at an angle from the ground at a speed of 30 m/s.

So, this motion will be projectile motion.

(2). We need to calculate the vertical and horizontal components of the initial velocity of the ball

Using formula of vertical initial velocity

`u_(y)=u\cos\theta`

Put the value into the formula

`u_(y)=30\cos40`

`u_(y)=22.9\ m/s`

Horizontal initial velocity ,

`u_(x)=u\sin\theta`

Put the value into the formula

`u_(x)=30\sin40`

`u_(x)=19.3\ m/s`

(3). We need to calculate the time

Using formula of time

`t=(2u\sin\theta)/(g)`

Put the value into the formula

`t=(2*19.3)/(9.8)`

`t=3.9\ sec`

(4). We need to calculate the range of the ball assuming the ground is flat and horizontal

Using formula of range

`R=(v^2\sin2\theta)/(g)`

Put the value into the formula

`R=(30^2\sin80)/(9.8)`

`R=90.4\ m`

Hence, (1). This motion is projectile motion.

(2). The horizontal initial velocity is 19.3 m/s.

The vertical initial velocity is 22.9 m/s.

(3). The time is 3.9 sec.

(4). The range of the ball assuming the ground is flat and horizontal is 90.4 m.