Question

A softball needs to hit a bucket43.0 m away on flat ground. If itis aimed at 72.0°, what must itsinitial velocity be?

Answer 1

Given,

The range of the projectile, R=43.0 m

The angle at which the bucket is projected, θ=72.0°

The range of a projectile is the maximum displacement of the projectile in the x-direction.

And the range is given by,

`R=(u^2\sin (2\theta))/(g)`

Where u is the initial velocity and g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} 43.0=(u^2*\sin (2*72.0^(\circ)))/(9.8) \\ u=\sqrt{(43*9.8)/(\sin(2*72.0^(\circ)))} \\ =26.77\text{ m/s} \end{gathered}`

Thus the initial velocity must be 26.77 m/s