Question

1. find the max. horizontal range of a cricket ball is
projected with a velocity of 3.92 m/s If the ball is
to have a range of 78°45√3 m. find the least angle
of projection and least time taken (Take g = 9.8 m/s 3)​

Answer 1

Answer:

u = 80m/s

R=100*3-1/2

R = (u2 Sin2b) / g

100*3-1/2 = 80*80 * Sin2b / 10

5*3-1/2/32 = Sin2b

sun2b = 0.67

2b = 42deg

b=21degree

Answer 2

Answer:

Explanation:

The ball is projected obliquely so suppose of projection from horizontal was θ

Then the horizontal range should be R=  

g

u  

2

Sin2θ

​  

 

As given in the question R=245m and u=49m/s

putting above values in the expression of R we get Sin2θ=1 so 2θ=90  

0

 or θ=45  

0

 so Sinθ=  

2

 

2

​  

 

1

​  

 

So the time in air or the time of flight will be T=  

g

2uSinθ

​  

=  

9.8

2×49Sinθ

​  

=5  

2

 

2

​  

Sec