Question

A player kicks a football obliquely at a speed of 20m/s so that its range is maximum. another player at a distance of 24m away in the direction of kick starts running at thet instant to catvh the ball. before the ball hits th ground to each it, the speed with speed with which the the second player has to run iss g=10

Answer 1

For maximum horizontal range,

Angle of projection (θ) = 45°

Speed of projection (u) = 20 m/s

Maximum Horizontal Range,

$\bf \implies R_{max} = \dfrac{u^2 sin 2 \theta}{g} \\ \\ \rm \implies R_{max} = \dfrac{20^2 \times sin 90^{\circ}}{10} \\ \\ \rm \implies R_{max} = \dfrac{40\cancel{0} \times 1}{\cancel{10}} \\ \\ \rm \implies R_{max} = 40 \ m$

Time of flight,

$\bf \implies T = \dfrac{2u sin \theta}{g} \\ \\ \rm \implies T = \dfrac{2 \times 2\cancel{0} \times sin 45^{\circ}}{\cancel{10}} \\ \\ \rm \implies T = 4 \times \dfrac{1}{\sqrt{2}} \\ \\ \rm \implies T = 2\sqrt{2} \ s$

So,

The distance that second player needs to run to catch ball in $\rm 2\sqrt{2} \ s$ = A - B

= 40 - 24

= 16 m

If 'v' is the speed of the player. Then,

$\bf \implies Speed = \dfrac{Distance \ travelled}{Time \ taken} \\ \\ \rm \implies v = \dfrac{16}{2\sqrt{2}} \\ \\ \rm \implies v = 4\sqrt{2} \ ms^{-1}$

$\therefore$ $\boxed{\mathfrak{Speed \ of \ second \ player \ (v) = 4\sqrt{2} \ m/s}}$