91Ó°ÊÓ

From the graph we can say that maximum range of projectile 240 m is which can be reaching by it only when launching angle is$45°$.

First, find the time to reach the maximum height for the object. We know that at the maximum height the object has only horizontal velocity component. Its vertical will be zero. So,the angle for half of maximum height can be found. After that from the graph we can conclude the initial velocity of object.

Formulae:

Time of flight,

$t=\frac{2v_0\sin {\mathrm{θ}}_0}{g}$ …(¾±)

Range of the projectile,

$R=\frac{v_0^2\sin 2\mathrm{θ}}{g}$ …(¾±¾±)

The x component of the velocity,

$v_{ox}=v_o\cos \mathrm{θ}$ …(¾±¾±¾±)

We know that that time to reach the maximum height is

$t=\frac{2v_0\sin {\mathrm{θ}}_0}{g}$

Time required to reach the maximum height by the object when$\mathrm{θ}=90°$is,

$t_{maximum}=\frac{2v_0}{g}$

Therefore, half of the maximum time of flight is given by,

$\frac{t_{maximum}}{2}=\frac{v_0}{g}$

That means,

$$\begin{gathered} \sin {\mathrm{θ}}_0=\frac{1}{2} \\ {\mathrm{θ}}_0=30° \end{gathered}$$

We know from graph is range is 240 m, and angle is $45°$. So, we can find its initial velocity.

$R=\frac{v_0^2\sin 2\mathrm{θ}}{g}$

After plugging values, we get$v_0=48.5m/s$

And at maximum height object has only horizontal velocity component

$v_{ox}=v_0\cos \mathrm{θ}$

So, at half maximum height object have velocity will,

$$\begin{gathered} v_{ox}=(48.5m/s)\cos (30°) \\ =42.0m/s \end{gathered}$$

Therefore, the least speed of the ball at given condition.