Question

A boy throws a table tennis ball of mass 200g
upwards with a velocity of 10 m/s at an angle $
with the vertical. The wind imparts a horizontal
force of 0.08 N, so that the ball returns to a
starting point. Then, the angle $ must be suchen
tan$ is
(1) 0.2
(2) 0.4
(3) 2.5
(4) 1.2​

Answer 1

Answer:

https://www.askiitians.com/forums/General-Physics/9/24256/motion.htm

Uy = U0COS@ & ay = -g

applying VY = UY + ayt

Vy at topmost point is zero so

t= uy/g

tim of flight (T)= 2t=2Uy/g .........1

for horizontal motion

Ux = UOCOS@

Sx = UxT + axT2

finally particle reaches to same point so displacement is zero ..

0 = UxT + axT2/2 .........2

putting value of T in eq 2

Uy/Ux = g/a .........3

max = force =0.08

ax = 4m/s2

now , UY/UX = Uocos@/Uosin@

=tan@=10/4=2.5

option c is correct