Question

A ball is kicked at 30° to the horizontal at 4.5m/s.calculate the range

Answer 1

Approximately
`1.8\; {\rm m}`, assuming that air resistance is negligible and
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

The range of a projectile (such as the ball in this question) is the same as the horizontal displacement between where the projectile was launched and where it landed.

While the ball is in the air, the horizontal component of velocity would be constant. At the same time, the vertical component of velocity would change with a constant acceleration of
`a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`.

The range of this ball can be found in the following steps:

• Find horizontal and vertical components of initial velocity.
• Find the duration of the flight.
• Using the fact that horizontal velocity is constant, find the range (distance travelled in the horizontal direction) by multiplying horizontal component of velocity by the duration of the flight.

Let
`u = 4.5\; {\rm m\cdot s^(-1)}` denote initial velocity. At
`\theta = 30^(\circ)` above the horizon:

• Horizontal component of initial velocity would be
  `u_(x) = u\, \cos(\theta)`.
• Vertical component of initial velocity would be
  `u_(y) = u\, \cos(\theta)`.

Assume that air resistance is negligible. By the conservation of energy, the vertical component of velocity right before landing,
`v_(y)`, would have the same magnitude as the vertical component of the initial velocity,
`u_(y)`. However, these two velocities would be opposite in direction:

`v_(y) = (-u_(y))`.

The change in velocity would be
`v_(y) - u_(y) = (-u_(y)) - u_(y) = (-2)\, u_(y)`.

Divide the change in the vertical component of velocity by the vertical component of acceleration to find the duration
`t` of the motion:

`\begin{aligned}t &= (v_(y) - u_(y))/(a_(y)) \\ &= ((-u_(y)) - \, u_(y))/((-g)) \\ &= (2\, u_(y))/(g)\end{aligned}`.

The horizontal component of velocity
`v_(x)` is constant under the assumption that air resistance is negligible. The value of the horizontal component of velocity would be equal to the initial value
`u_(x)`. The range
`x_(x)` (horizontal displacement) of the ball would be equal to the product of this horizontal component of velocity and the duration of the flight:

`\begin{aligned}x_(x) &= u_(x)\, t \\ &= (u\, \cos(\theta))\, \left((2\, u_(y))/(g)\right)\\ &= (u\, \cos(\theta))\, \left((2\, u\, \sin(\theta))/(g)\right) \\ &= (u^(2)\, (2\, \sin(\theta)\, \cos(\theta)))/(g)\end{aligned}`.

Using the trigonometric identity
`2\, \sin(\theta)\, \cos(\theta) = \sin(2\, \theta)`, the expression for range can be further simplified:

`\begin{aligned}x_(x) &= u_(x)\, t \\ & \cdots \\ &= (u^(2)\, (2\, \sin(\theta)\, \cos(\theta)))/(g) \\ &= (u^(2)\, \sin(2\, \theta))/(g)\end{aligned}`.

Substitute
`u = 4.5\; {\rm m\cdot s^(-1)}`,
`\theta = 30^(\circ)`, and
`g = 9.81\; {\rm m\cdot s^(-2)}` into this expression to obtain:

`\begin{aligned}x_(x) &= (u^(2)\, \sin(2\, \theta))/(g) \\ &= ((4.5)^(2)\, \sin(2\, (30^(\circ))))/(9.81)\; {\rm m} \\ &\approx 1.79\; {\rm m}\end{aligned}`.

In other words, the range of this ball would be approximately
`1.8\; {\rm m}` under the assumptions (rounded to two significant figures.)