Question

A ball is thrown from ground level so as to just clear
a wall 4 metres high at a distance of 4 m. and falls at
a distance of 14 m. from the wall, then the magnitude
of the velocity of the ball is-

Answer 1

Answer:

in this case the magnitude is 10 m

Answer 2

Answer:
$\sqrt{182}$

Explanation:

The ball passes through point P(4,4).
x=4 and y=4
Also, its range=4+14=18m.

The trajectory of the ball is,

y=xtan∅[1-$\frac{x}{R}$]

substituting the equation we get:
4=4tan∅×$\frac{7}{9}$
therefore; ∅=tan⁻¹($\frac{9}{7}$)

R= $\frac{2u^{2} sinthetacostheta}{g}$
18=$\frac{2}{9.8}$×u²×$\frac{9}{\sqrt[2]{130} }$×$\frac{7}{\sqrt{130} }$

u=$\sqrt{182}$