Question

If a baseball player throws a ball at 35.0 m/s what is its maximum range?

Answer 1

`124.87m`, rounded to two decimal places.

Explanation:

The maximum horizontal distance traveled by the base ball is the horizontal distance it covers when it returns to its initial height. Let it be `(y = 0)` as shown below.
If the ball is thrown making an angle `theta` with the horizontal with velocity `v_0`
Using the kinematic equation
`s=ut+1/2g t^2`,

Noting that acceleration due to gravity `g=9.81ms^-2` is opposing the direction of motion. we get for vertical component of velocity
`0=v_{0}t\sintheta -\frac {1}{2}g t^{2}`
`=>0=v_{0}\sintheta -\frac {1}{2}g t`

Time `t` taken to return to ground or to travel distance `d` is found as
`t = (2 v_0 sin ⁡ θ) / g` ....(1)

Ignoring air resistance,
Distance `d="Horizontal component of velocity"xx"time"`
`=(v_{0}cos\theta )t` .....(2)

Inserting value of `t` from (1) in (2) we get
`d=(v_{0}cos\theta )(2 v_0 sin ⁡ θ) / g`
`=>d=v_{0}^2/g 2sin theta cos\theta`
`=>d=v_{0}^2/g sin 2theta`
For its maximum range `sin 2 theta=1`
`=>d_max=v_{0}^2/g`

Taking Inserting values we get
`d_max=(35.0)^2/9.81`
`=>d_max=124.87m`, rounded to two decimal places.