Question

the ratio of (a) maximum heights attained by them and (b) horizontal
A ball is thrown at angle 8 and another ball is thrown at angle (90° - 0) with the horizontal direction from the
same point each with speeds of 40 m/s. The second ball reaches 50m higher than the first ball. Find then
individual heights. g = 10 m/s2
Thore of a particle when launched at an angle of 15° with the horizontal is 1.5 km. What is the range of the​

Answer 1

Answer:

u = 40 m/s

H  

2

​  

=H  

1

​  

+50                     ...(1)

Here H  

1

​  

 = Height reached by first ball  

H  

2

​  

= Maximum height reached by second ball  

g=9.8ms  

−2

 

H  

1

​  

=  

2g

4  

2

sin  

2

θ

​  

                                      H  

2

​  

=  

2g

4  

2

sin  

2

(90−θ)

​  

 

H  

1

​  

=  

19.6

(40)  

2

sin  

2

θ

​  

                               H  

2

​  

=  

19.6

(40)  

2

sin  

2

(90−θ)

​  

 

Putting value of (2nd) in (1st)

H  

1

​  

=  

19.6

(40)  

2

cos  

2

θ

​  

   =    H  

1

​  

=  

19.6

(40)  

2

sin  

2

θ

​  

+50

by solving  

H  

1

​  

=20m

H  

2

​  

=70m

THINK SO THIS IS THE ANSWER

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