Question

the horizontal range of a projectile fired at an angle of 15° is 50 m .it is fired with with the same speed at an angle of 45 degree its range will be what​

Answer 1

Answer:

Explanation:

INITIAL CONDITION  :

thetha = 15 degree

range = 50m

LET INITIAL SPEED BE u .

THEREFORE,

range = u^2   ×  sin[2thetha]    /  g

50  =  u^2 × sin30 /  10           {g=10}

500 ×  2    = u^2

u  =  √[1000]

u = 10√10

FINAL CONDITION  :

thetha = 45 degree

u = 10√10

range = [10√10]^2  × sin90/10

range = 1000 × 1 /10

range = 100m

HERE IS YOUR ANSWER,

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