Question

During the flight of the ball to the wicketkeeper, the horizontal velocity remains unchanged. The speed at the moment the wicketkeeper catches it is 19ms calculate its vertical speed

Answer 1

Answer:

ANSWER

Given : u=28 m/s θ=30

o

(a) : Maximum height H=

2g

u

2

sin

2

θ

​

⟹ H=

2×9.8

28

2

×0.5

2

​

=10 m

(b) : Time of flight T=

g

2usinθ

​

⟹ T=

9.8

2×28×sin30

​

=2.86 s

(c) : Range R=ucosθ×T

⟹ R=28×cos30

o

×2.86=69.27 m