COMEDK 2024 Shift 2 Paper

To determine the maximum height attained by the ball, we can use the equations of projectile motion. Given the initial height of the ball is same as the height of the cricketers (both are 2.5 m tall), the angle of projection is 30∘, and the horizontal range is 50 m , we need to find out the maximum height.
First, let's calculate the initial velocity of the ball. For horizontal range, we use the formula:
R=‌

u2sin‌2θ

g

where:
R is the range =50m
u is the initial velocity
θ is the angle of projection =30∘
g is the acceleration due to gravity =9.8m∕ s2
First, calculate sin‌2θ :
sin‌2θ=sin‌60∘=‌

√3

2

Thus, the range equation becomes:
50=‌

u2⋅ √3 2

9.8

Solve for the initial velocity u :
‌u2=‌

50⋅9.8⋅2

√3

‌u2=‌

980

√3

‌u≈√565.69
‌u≈23.78m∕ s
Now, we need to find the maximum height achieved by the ball. The formula for maximum height in projectile motion is:
H=h0+‌

u2sin‌2θ

2g

where:
h0 is the initial height =2.5m
sin‌θ is for θ=30∘
First, calculate sin‌30∘ :
sin‌30∘=0.5
Thus, substituting the values, we get:
‌H=2.5+‌

23⋅782⋅0.52

2⋅9.8

‌H=2.5+‌

23.782⋅0.25

19.6

‌H=2.5+‌

565⋅6⋅⋅0.25

19⋅6

‌H=2.5+‌

141.42

19.6

‌H≈2.5+7.2
‌H≈9.7m
Therefore, the maximum height attained by the ball is closest to option D, which is:
Option D: 9.7 m