Question

During the flight of the ball to the wicket-keeper, the horizontal velocity remains unchanged. The speed of the ball at the moment the wicket-keeper catches it is 19 m/s. Calculate for the ball just as it is caught:

i) Its vertical speed,

ii) The angle that the path of the ball makes with the horizontal.

iii) Suggest with a reason whether the ball, at the moment it is caught, is rising or falling.

a. i) 0 m/s, ii) 90 degrees, iii) Falling

b. i) 19 m/s, ii) 0 degrees, iii) Rising

c. i) 9.5 m/s, ii) 45 degrees, iii) Rising

d. i) 19 m/s, ii) 90 degrees, iii) Falling

Answer 1

The vertical speed of the ball is 0 m/s, the angle is 90 degrees, and the ball is falling when caught. Therefore correct option is a. i) 0 m/s, ii) 90 degrees, iii) Falling.

Step-by-step explanation:

To calculate the vertical speed and angle of the ball just as it is caught, we need to analyze the projectile motion of the ball. In projectile motion, the horizontal velocity remains unchanged throughout the flight. Given that the speed of the ball at the moment it is caught is 19 m/s, we can determine the vertical speed and angle:

1. The vertical speed of the ball is 0 m/s because at the moment of catch, the ball has reached its peak height and is momentarily at rest before falling back down.
2. The angle that the path of the ball makes with the horizontal is 90 degrees because when the ball is at its peak height and about to fall, its path is perpendicular to the horizontal.
3. The ball, at the moment it is caught, is falling because its vertical speed is 0 m/s and it has completed its upward trajectory and is about to descend.