AP EAMCET202423 May 2024Morning ShiftPhysicsMotion in Two DimensionsActual
Path of projectile is given by the equation Y=P x-Q x^2 , match the following accordingly (acceleration due to gravity = g) array |l|l|l|l| A & Range & (i) & P Q B & Maximum height & (ii) & P C & Time of flight & (iii) & P^2 4 Q D & Tangent of projection & (iv) & ( 2 g Q ) P array
Options
- AA-i,B-iii, C-iv, D-ii
- BA-i,B-iii, C-ii, D-iv
- CA-iii,B-i, C-iv, D-ii
- DA-iv,B-ii, C-iii, D-i
Correct answer
A. A-i,B-iii, C-iv, D-ii
Step-by-step solution
Path of projectile, Y = Px - Q x ^2 array ll & At x=R, Y=0 & 0=P R-Q R^2 R(P-Q R)=0 array Range, R = P Q At x = R 2 , Y = H Maximum Height, H = P ( R 2 )- Q ( R 2 )^2 aligned & =P ( P 2 Q )-Q ( P 2 Q )^2 & = P^2 2 Q - P^2 4 Q = P^2 4 Q aligned Tangent of projection, = P Also, g 2 u ^2 ^2 = Q u = g 2 Q R = T ( ucos ) T = P Q ~g 2 Q = ( 2 gQ ) P