SATHEE: Kinematics Question 691

Kinematics Question 691

Question: A cricketer his a ball with a velocity $25 m / s$ at $60^{o}$ 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)

Options:

A) 8.2 m

B) 9.0 m

C) 11.6 m

D) 12.7 m

Show Answer

Answer:

Correct Answer: A

Solution:

[a]Horizontal component of velocity $v_{x} = 25 \cos 60^{\circ} = 12.5 m / s$

Vertical component of velocity $v_{y} = 25 \sin 60^{\circ} = 12.5 \sqrt{3} m / s$

Time to cover 50 m distance $t = \frac{50}{12.5} = 4 \sec$

The vertical height y it’s given by

$y = v_{y} t - \frac{1}{2} g t^{2} = 12.5 \sqrt{3} \times 4 - \frac{1}{2} \times 9.8 \times 16 = 8.2 m$

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Kinematics Question 691

Question: A cricketer his a ball with a velocity 25m/s at 60o 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)

Options:

Answer:

Solution:

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Question: A cricketer his a ball with a velocity $25 m / s$ at $60^{o}$ 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)