Question

From the ground level, a ball is to be shot with certain speed. graph shows range R it will have verses the launch angle thetha . the least speed the ball will have during its flight if thetha is choosen such that the flight time is half of its maximum possible value, is equal to...

Answer 1

Final Answer : $25 \sqrt{3} m/s$

Symbols has usual meanings.
Take g = 10m/s^2

Steps:
1) Main Points: Initial velocity of Projection is same in all cases, regardless of $\theta$ .
So, we will find initial velocity by taking $\theta$ which suits us.

2) We will take $\theta = 45 \degree$ where Range is maximum (250m) .
$R_{max}= \frac{u^2}{g} \\ => 250 m = \frac{u^2}{10} \\ => u = 50 m/s .$

3) According to the question,
$T = \frac{1}{2} T_{max} \\ \frac{2u \sin(\theta) }{g} = \frac{1}{2} *\frac{2u }{g } \\ => \theta = 30\degree$

4) Least speed of ball during it's time of motion , when it reaches it's maximum height and there vertical component of velocity gets 0 .
Remains is Horizontal Component :
$v = u \cos(\theta ) \\ => v = 50 * \cos(30\degree) \\ => v = 50 *\frac{\sqrt{3}}{2} \\ => v = 25 \sqrt{3} m/s$

Hence, Required speed is $25 \sqrt{3} m/s$ .

Answer 2

Range maximum =sin90degree

R =u^²/g

250=u^²/g

U=50

Time maximum=2usin thetha/g

1/2=2u/g

Thetha = 30°

We have formula

V=using thetha

Sin30°=√3/2

50*√3/2

Answer

25√3