Question

cricketer throws a ball at an angle of 50 degree with initial velocity 90 kilometre per hour find the maximum height​

Answer 1

Answer:

50m

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