A cricketer hits a ball with a velocity 25 m/s at 60° above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground) :

Question

A cricketer hits a ball with a velocity 25 m/s at 60° above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground) : (A) 8.2 m (B) 9.0 m (C) 11.6 m (D) 12.7 m

Tardigrade Admin · Accepted Answer

Here ball is hit by cricketer with 25 m/s velocity. The angle of projection

= 60^{o}

Horizontal component of velocity

υ_{x} = 25 cos 60^{o}

= 12.5 m / s

Vertical component of velocity

υ_{y} = 25 sin 60^{o}

= 25 \times \frac{3}{2} = 12.5 3 m / s

Time taken by ball to cover 50 m distance

t = \frac{s}{υ _{x}} = \frac{50}{12.5} = 4 sec

Now the vertical distance covered by the ball is given by

x = υ_{y} t - \frac{1}{2} g t^{2}

= 12.5 3 \times 4 - \frac{1}{2} \times 9.8 \times (4)^{2}

= 503 - 78.4

86.6 - 78.4

= 8.2 m

Q. A cricketer hits a ball with a velocity 25 m/s at 60∘ above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground) :

8.2 m

9.0 m

11.6 m

12.7 m

Q. A cricketer hits a ball with a velocity 25 m/s at $60^{\circ}$ above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground) :