Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of launch is 75.0° and 15.0° respectively. Which statement is true for the two projectiles?
Answers
Answered by
3
Your question is incomplete ,
A complete question is -----> Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of the launch is 75.0 degrees and 15.0 degrees respectively. Which statement is true for the two projectiles?
A) The range of A and B are equal
B) The height of A is less than the height of B
C) The height of A and B is equal
D) The range of A is more than the range of B
E) The range of B is more than the range of A.
solution :- you know, range = u²sin2θ/g
Now, in case 1 :- θ = 75° , intial velocity = u ( assume )
Then, Range = u²sin2(75°)/g
= u²sin150°/g
= u²/2g
in case 2 :- θ = 15° , intial velocity = u { according to question , velocities are same}
Then, range = u²sin2(15°)/g = u²sin30°/g
= u²/2g
Hence, range of A and B are equal .
You also know, Height = u²sin²θ/2g
in case 1 :- θ = 75° and initial velocity = u
Height = u²sin²75°/2g
In case 2 :- θ = 15° , intial velocity = u
height = u²sin²15°/2g
Here we observed , height of A is greater than height of B , because sin²75°>sin²15°
Hence,only option (A) is correct
A complete question is -----> Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of the launch is 75.0 degrees and 15.0 degrees respectively. Which statement is true for the two projectiles?
A) The range of A and B are equal
B) The height of A is less than the height of B
C) The height of A and B is equal
D) The range of A is more than the range of B
E) The range of B is more than the range of A.
solution :- you know, range = u²sin2θ/g
Now, in case 1 :- θ = 75° , intial velocity = u ( assume )
Then, Range = u²sin2(75°)/g
= u²sin150°/g
= u²/2g
in case 2 :- θ = 15° , intial velocity = u { according to question , velocities are same}
Then, range = u²sin2(15°)/g = u²sin30°/g
= u²/2g
Hence, range of A and B are equal .
You also know, Height = u²sin²θ/2g
in case 1 :- θ = 75° and initial velocity = u
Height = u²sin²75°/2g
In case 2 :- θ = 15° , intial velocity = u
height = u²sin²15°/2g
Here we observed , height of A is greater than height of B , because sin²75°>sin²15°
Hence,only option (A) is correct
Similar questions