Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Rest and Motion Kinematics
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rest and Motion Kinematics
The benches of a gallery in a cricket stadium are 1 m wide and 1 m...
The benches of a gallery in a cricket stadium are wide and high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at at an angle of 53^{}^{{}^{\circ}} with the horizontal. The benches are perpendicular to the plane of motion and the first bench is from the batsman. Determine the number of the bench on which the ball will land. (Provide the bench number as an integer).
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
The total horizontal distance the bike must cover to safely cross the ditch is the sum of the width of the ditch and the horizontal length of the bike.
Horizontal length of the bike = L \cos \theta = 5 \cos 15^{}^{{}^{\circ}} \approx 5 \times 0.966 = 4.83 ft.
Total required horizontal range ft.
Using the formula for the horizontal range of a projectile, .
Substituting ft, \theta = 15^{}^{{}^{\circ}}, and standard gravity in US customary units ft/s^216.53 = \dfrac{v^2 \sin(30^{}^{{}^{\circ}})}{32}16.53 = \dfrac{v^2 \times 0.5}{32}v^2 = \dfrac{16.53 \times 32}{0.5} = 1057.92v = \sqrt{1057.92} \approx 32.5 ft/s.
Horizontal length of the bike = L \cos \theta = 5 \cos 15^{}^{{}^{\circ}} \approx 5 \times 0.966 = 4.83 ft.
Total required horizontal range ft.
Using the formula for the horizontal range of a projectile, .
Substituting ft, \theta = 15^{}^{{}^{\circ}}, and standard gravity in US customary units ft/s^216.53 = \dfrac{v^2 \sin(30^{}^{{}^{\circ}})}{32}16.53 = \dfrac{v^2 \times 0.5}{32}v^2 = \dfrac{16.53 \times 32}{0.5} = 1057.92v = \sqrt{1057.92} \approx 32.5 ft/s.