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A cricketer hits a ball with a velocity25 \ 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-)

Option: 1

8.2m


Option: 2

9.0m


Option: 3

11.6 m


Option: 4

12.7 m


Answers (1)

best_answer

Horizontal component of Velocity 
Vx =25\ cos \ 60^{\circ}= 12.5\ m/s
Vertical component of Velocity 
Vy = 25\ sin\ 60^{\circ}= 12.5\sqrt{3}\ m/s

Time to cover 50 m distance t = \frac{50}{12.5}=\frac{500}{125}
                                              t = 4\ sec
The vertical height y is given by- 
\begin{aligned} & y=V_y t-\frac{1}{2} g t^2 \\ & y=12.5 \sqrt{3} \times 4-\frac{1}{2} \times 9.8 \times 16 \\ & y=50 \sqrt{3}-9.8 \times 8=50 \times 1.732-9.8 \times 8 \\ & y=86600-78.4 \end{aligned}
y=86.6-78.4=8.2 \mathrm{~m}

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