Question

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Answer it quick!!!!!
The equations of motion of a projectile thrown in x–y plane from origin are x = 8t, y = 6t – 10t2
then the
angle of projectile is ????

Answer 1

37 degrees is answer.first differentiate it then velocity x component =8 and velocity y component=6-10t.then keep t=0.then do 6/8=tantheta .theta=37

Answer 2

Answer: 37°

Given data says that motion in x-y plane has two equation such as x = 8t and $y= 6t-10 t^2$.

To get the angle of projectile, we just need to calculate the velocity in the x – y dimension that is, x component of velocity and y component of velocity. Therefore differentiate the two equations and we get,

                      $-v_x = dx / dt = dx / dt( 8t) = 8$

and for,

                      $v_y = dy / dt = dy / dt ( 6t-10 t ^2) = 6-10t.$

Assume t = 0 and then form the equation in terms of velocity,

$v = v_y / v_x\\= v sin \theta /v cos \theta \\= 6 / 8$

=> $tan \theta = 3 / 4$

=> $\theta = 37 degrees.$