What is the angle of projection with given initial velocities?

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To calculate the angle of projection when the initial velocity at the highest point is half of the initial velocity, one must recognize that the vertical component of velocity is zero at this point, while the horizontal component remains unchanged. The relationship between the initial horizontal velocity and the initial speed is crucial, as it can be expressed using the equation v_0^2 = v_x^2 + v_y^2. Given that v_y is zero at the highest point, it follows that v_x equals half of the initial speed. The discussion emphasizes understanding the dynamics of projectile motion, particularly the distinction between vertical and horizontal components. Clarification on these concepts is essential for solving the problem accurately.
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help me pls
if the initial velocity of a projectile at the highest point of trajectory is equal to half of the initial velocity of projection, calculate the angle of projection.


can anyone help me
 
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First of all, say "speed" rather than velocity here!

Furthermore:
1) What is the vertical velocity component at the highest point of the trajectory?
2) Does the horizontal velocity component change during projectile motion?
(I have assumed that air resistance is negligible)
3) How is the initial horizontal velocity component related to the initial speed and angle of projection?

Try to formulate answers to these questions.
 
it is not speed.
sorry for the late reply as i had exams. This question is from my physics question paper. the question is as i have written. no more info was give. only g acts here. note that at the higest point of a projectile the vertical v=0 but the horizontal velocity remain the same unaltered.
 
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v_0^2 = v_x^2 + v_y^2

When v_y^2 = 0, v_x = \frac{1}{2} v_0 [/tex]<br /> <br /> Draw yourself a triangle.
 
whozum said:
v_0^2 = v_x^2 + v_y^2

When v_y^2 = 0, v_x = \frac{1}{2} v_0 [/tex]<br /> <br /> Draw yourself a triangle.
<br /> <br /> v_m_a_x_h = 0<br /> v_i_n_t_i = 1/2 v_m_a_x [/tex]&lt;br /&gt; v max= 0 therefore 1/2 of 0 = 0 whis not the correct answer.
 
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at h = hmax, vy = 0, vx however is not zero.
 
therefore u are telling that
v_m_a_x = 1/2 v_x
which implies that
v_m_a_x = 1/2 u cos x [/tex]<br /> i think u are right.<br /> i made a mistakwe here only.<br /> i knew all other questions but this one confused me a lot. anyway thank u
 
Do you understand WHY all this happens though?
 
no
it just didnt strike me.
moreover i was not there when sir explained this sum before the exam
 
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  • #10
So you do understand, or you dont?
Would you like me to explain why?
 
  • #11
i understand,
but i didnt understand the 2nd part of ur question
 
  • #12
Would you like me to explain anything about your problem.

Fhimta'l Su'al or no? As3dak?
 
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