Question

A cricketer can throw a ball to a maximum horizontal distance of 100 m with the same speed how high above the ground can be cricketer throw the ball​

Answer 1

Step-by-step explanation:Maximum horizontal distance, R=100m  

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45  

∘

, i.e.,  =45  

∘

.

The max horizontal range for a projection velocity v is given by the relation:

R  

max

​  

=  

g

u  

2

 

​  

 

100=  

g

u  

2

 

​  

 

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.

Acceleration, a=g

Using the third equation of motion:

v  

2

−u  

2

=−2gH

H=u  

2

/2g=100/2=50m

Answer 2

Step-by-step explanation:

h=100m

r=u^2/gi 12

100=u^2/10

:u=10√10

h= u^2/g2

h=1000/20

h=50